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Start with a locally-Minkowski basis with metric $(g')_{ij} = \eta_{ij}$.
Reminder that the Minkowski metric is defined as $\eta_{ij} = \delta_{ij} \cdot \sigma_i$ for $\sigma_i = \pm 1$.
Let its signature be $\sigma = \overset{n}{\underset{i=1}{\Pi}} \sigma_i = det[\eta_{ij}] = \pm 1$.

Let's give it a transform from our choice of (not necessarily coordinate) bais:
$g_{ab} = {(e')_a}^i {(e')_b}^j (g')_{ij} = {(e')_a}^i {(e')_b}^j \eta_{ij}$

Let's define the Levi-Civita permutation tensor to be a tensor in the basis of the Minkowski metric:
$(\epsilon')_{i_1 ... i_n} = 1$ for even permutations of $i_1 ... i_n$, -1 for odd permutations, and 0 otherwise.

The contravariant-form representation of this tensor is:
$(\epsilon')^{i_1 ... i_n} = \eta^{i_1 j_1} ... \eta^{i_n j_n} (\epsilon')_{j_1 ... j_n} = det([\eta^{ij}]) (\epsilon')_{i_1 ... i_j} = \sigma^{-1} (\epsilon')_{i_1 ... i_n} = \sigma (\epsilon')_{i_1 ... i_n} $.

Then the same tensor in our (not necessarily coordinate) basis is:
$\epsilon_{a_1 ... a_n} = {(e')_{a_1}}^1 ... {(e')_{a_n}}^n (\epsilon')_{1 ... n}$, which is the definition of $det([{(e')_a}^i])$, which is the linear transform from the Minkowski-metric basis to the new basis.

So before we had:
$\tilde{g}_{\tilde{a}\tilde{b}} = {e^a}_\tilde{a} {e^b}_\tilde{b} g_{ab}$.
with determinant:
$\tilde{g} = e^2 g$
Now we are saying $g_{ab} = {(e')_a}^i {(e')_b}^j \eta_{ij}$.
with determinant:
$g = (e')^2 \sigma$, for $e' = det([{(e')_a}^i])$.
So combine to find: $\tilde{g}_{\tilde{a}\tilde{b}} = {e^a}_\tilde{a} {e^b}_\tilde{b} {(e')_a}^i {(e')_b}^j \eta_{ij}$.
and determinant:
$\tilde{g} = e^2 (e')^2 \sigma$
Solve for $e'$ to find:
$e' = \frac{\sqrt{\tilde{g}}}{e} = \sqrt{g}$

So the determinant of the transform from the Minkowski metric to the non-coordinate basis is $det([{(e')_a}^i]) = \sqrt{g}$.
Therefore $ \epsilon_{a_1 ... a_n} = e' = \sqrt{|g|}$ for even permutations, $-e' = -\sqrt{|g|}$ for odd permutations, and 0 otherwise.

$\epsilon^{a_1 ... a_n} = \sigma (g^{a_1 b_1}) \cdot ... \cdot (g^{a_n b_n}) \epsilon_{b_1 ... b_n}$
$= \sigma \frac{1}{|g|} \epsilon_{a_1 ... a_n}$
$= \sigma \frac{1}{\sqrt{|g|}}$ when $a_1 ... a_n$ is even, negative this when it is odd.

Levi-Civita tensor for 4D Lorentzian signature with a diagonalized metric:
$\epsilon^{ijkl} = -\epsilon_{ijkl}$
the timelike (zero) index goes first, so $\epsilon_{0123} = 1$

Levi-Civita tensor in 4D manifold with Lorentzian signature:
$\epsilon_{txyz} = (-g)^{\frac{1}{2}}$
$\epsilon^{txyz} = -(-g)^{-\frac{1}{2}}$

generalized Kronecker delta:
$\delta^{i_1 ... i_n}_{j_1 ... j_n} = n! \delta^{i_1}_{[j_1} \cdot ... \cdot \delta^{i_n}_{j_n]} = \sigma \epsilon^{i_1 ... i_n} \epsilon_{j_1 ... j_n}$

contractions of Levi-Civita permutation tensor:
$\epsilon_{i_1 ... i_n} \epsilon^{j_1 ... j_n} = \sigma \delta^{j_1 ... j_n}_{i_1 ... i_n}$
Should I define the sub-rank generalized Kronecker delta in terms of the summation indexes at the front or the back of the Levi-Civita tensor? I think it doesn't matter.
$\epsilon_{i_1 ... i_k \ i_{k+1} ... i_n} \epsilon^{i_1 ... i_k \ j_{k+1} ... j_n} = \sigma \delta^{i_1 ... i_k \ j_{k+1} ... j_n}_{i_1 ... i_k \ i_{k+1} ... i_n} = \sigma k! \delta^{j_{k+1} ... j_n}_{i_{k+1} ... i_n}$
$\epsilon_{i_1 ... i_n} \epsilon^{i_1 ... i_n} = \sigma n!$

specific instances for 4D lorentzian space:
$\delta^{ijkl}_{mnpq} = -\epsilon^{ijkl} \epsilon_{mnpq}$
$\delta^{ijk}_{mnp} = \delta^{ijkl}_{mnpl} = -\epsilon^{ijkl} \epsilon_{mnpl}$
$2! \delta^{ij}_{mn} = \delta^{ijk}_{mnk} = \delta^{ijkl}_{mnkl} = -\epsilon^{ijkl} \epsilon_{mnkl}$
$3! \delta^i_j = 2! \delta^{ij}_{mj} = \delta^{ijk}_{mjk} = \delta^{ijkl}_{mjkl} = -\epsilon^{ijkl} \epsilon_{mjkl}$
$4! = 3! \delta^i_i = 2! \delta^{ij}_{ij} = \delta^{ijk}_{ijk} = \delta^{ijkl}_{ijkl} = -\epsilon^{ijkl} \epsilon_{ijkl}$

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Hodge Dual

Hodge dual of a k-form:
$\star (e^1 \wedge ... \wedge e^k) = {\epsilon^{1 ... k}}_{k+1 ... n} e^{k+1} \wedge ... \wedge e^n$
For arbitrary indexes:
$\star (e^{i_1} \wedge ... \wedge e^{i_k}) = \frac{1}{(n-k)!} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \wedge ... \wedge e^{i_n}$

The $\frac{1}{(n-k)!}$ is to cancel all the redundant sums of the wedge product, so this can also be written as:
$\star (e^{i_1} \wedge ... \wedge e^{i_k}) = {\epsilon^{i_1 ... i_k}}_{|i_{k+1} ... i_n|} e^{|i_{k+1}} \wedge ... \wedge e^{i_n|}$

As an outer product basis:
$k! \star (e^{[i_1} \otimes ... \otimes e^{i_k]}) = {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{[i_{k+1}} \otimes ... \otimes e^{i_n]}$
$\star (e^{[i_1} \otimes ... \otimes e^{i_k]}) = \frac{1}{k!} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{[i_{k+1}} \otimes ... \otimes e^{i_n]}$
If we consider the $\star$ operator to be antisymmetrizing on its basis then we don't need to worry about specifying the left hand side to be antisymmetric up front. And if we consider the Levi-Civita permutation tensor to be antisymmetrizing then we don't have to explicitly antisymmetrize the right hand side basis:
$\star (e^{i_1} \otimes ... \otimes e^{i_k}) = \frac{1}{k!} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \otimes ... \otimes e^{i_n}$
From there, product with tensor coefficients to get:
$\star (T_{i_1 ... i_k} e^{i_1} \otimes ... \otimes e^{i_k}) = \frac{1}{k!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \otimes ... \otimes e^{i_n}$
$\star T = \frac{1}{k!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \otimes ... \otimes e^{i_n}$

Therefore for 0-form 1:
$\star 1 = \frac{1}{n!} \epsilon_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n} = \epsilon_{1 ... n} e^1 \wedge ... \wedge e^n$
... using $\epsilon_{1...n} = \sqrt{|g|}$...
$= \sqrt{|g|} e^1 \wedge ... \wedge e^n$
Notice that only the dual of the 0-form does not involve any raising using the metric(-inverse).

Likewise for n-form:
$\star (e^1 \wedge ... \wedge e^n) = \frac{1}{0!} \epsilon^{1 ... n}$.
$= g^{1 i_1} \cdot ... \cdot g^{n i_n} \epsilon_{i_1 ... i_n}$
$= \frac{\sigma}{\sqrt{|g|}}$.

If the metric is diagonal then:
$\star (e^1 \wedge ... \wedge e^n) = \Pi_{j=1}^n g^{jj} \epsilon_{1 ... n} = \Pi_{j=1}^n g^{jj}$

Hodge dual of a k-form:
Define the Hodge dual on a k-form in n dimensions as:
Let $\star T = \star (T_{i_1 ... i_k} e^{i_1} \otimes ... \otimes e^{i_k})$
substitute the definition of wedge product basis for a (0 k) tensor:
$= \star ( \frac{1}{k!} T_{i_1 ... i_k} e^{i_1} \wedge ... \wedge e^{i_k})$
factor out the scalars and apply the Hodge dual to the wedge products only:
$= \frac{1}{k!} T_{i_1 ... i_k} \star ( e^{i_1} \wedge ... \wedge e^{i_k})$
insert in a factor of $\frac{1}{(n-k)!}$ since that's how many repeated elements we have due to the Levi-Civita permutation tensor:
$= \frac{1}{k!} \cdot \frac{1}{(n-k)!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \wedge ... \wedge e^{i_n}$
convert from wedge products to antisymmetric outer products, insert a $(n-k)!$ term to account fo the number of antisymmetric products.
$= (n-k)! \cdot \frac{1}{k!} \cdot \frac{1}{(n-k)!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{[i_{k+1}} \otimes ... \otimes e^{i_n]}$
factor out factorials, take the coefficient of the outer products of the form basis to be our componets of $(\star T)_{i_{k+1} ... i_n}$:
$= (\star T)_{i_{k+1} ... i_n} e^{i_{k+1}} \otimes ... \otimes e^{i_n}$
so for $T_{i_1 ... i_n}$ we have $(\star T)_{i_{k+1} ... i_n} = \frac{1}{k!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n}$

Hodge dual of a (0 k) tensor:
Lets generalize the Hodge dual definition to apply to any (0 k) tensor, to map to the forms by taking the antisymmetric component of the tensor:
$T = T_I e^I = T_{i_1 ... i_k} e^{i_1} \otimes ... \otimes e^{i_k}$
The antisymmetric part of it is equal to:
$T_{[I]} e^I = T_{[I]} e^{[I]} = \frac{1}{k!} T_{[I]} e^{\wedge I}$
$= \frac{1}{k!} T_{[i_1 ... i_k]} e^{i_1} \wedge ... \wedge e^{i_k}$
so ...
$(\star T)_{i_{k+1} ... i_n} e^{i_{k+1}} \wedge ... \wedge e^{i_n} = \frac{1}{k!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \wedge ... \wedge e^{i_n}$
$(\star T)_{i_{k+1} ... i_n} e^{i_{[k+1}} \otimes ... \otimes e^{i_n]} = \frac{1}{k!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{[k+1}} \otimes ... \otimes e^{i_n]}$
so for antisymetric $(\star T)_{i_{k+1} ... i_n}$ we find:
$(\star T)_{i_{k+1} ... i_n} e^{i_{k+1}} \otimes ... \otimes e^{i_n} = \frac{1}{k!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \otimes ... \otimes e^{i_n}$
so $(\star T)_{i_{k+1} ... i_n} = \frac{1}{k!} T_{i_1 ... i_k} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n}$

specific instances for 4D Lorentzian space:
For a 0-degree tensor: $\star T = T \epsilon_{abcd} dx^a \otimes dx^b \otimes dx^c \otimes dx^d$,
so $(\star T)_{abcd} = T \epsilon_{abcd}$
For a 1-degree tensor: $\star (T_a dx^a) = T_a {\epsilon^a}_{bcd} dx^b \otimes dx^c \otimes dx^d$,
so $(\star T)_{abc} = T_a {\epsilon^a}_{bcd}$
For a 2-degree tensor: $\star (T_{ab} dx^a \otimes dx^b) = \frac{1}{2} T_{uv} {\epsilon^{uv}}_{ab} dx^a \otimes dx^b = (\star T)_{ab} dx^a \otimes dx^b $,
so $(\star T)_{ab} = \frac{1}{2} T_{uv} {\epsilon^{uv}}_{ab}$.
For a 3-degree tensor: $\star (T_{abc} dx^a \otimes dx^b \otimes dx^c) = \frac{1}{3!} T_{abc} {\epsilon^{abc}}_d dx^d = (\star T)_d dx^d $,
so $(\star T)_d = \frac{1}{3!} T_{abc} {\epsilon^{abc}}_d$
For a 4-degree tensor: $\star (T_{abcd} dx^a \otimes dx^b \otimes dx^c \otimes dx^d) = \frac{1}{4!} T_{abcd} \epsilon^{abcd}$,
so $\star T = \frac{1}{4!} T_{abcd} \epsilon^{abcd}$

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Double Hodge Dual

Double Hodge dual star applied to a p-form:
$\star (\star (e^{i_1} \wedge ... \wedge e^{i_k}))$
$= \star ( \frac{1}{(n-k)!} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \wedge ... \wedge e^{i_n} )$
$= \frac{1}{(n-k)!} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} \star ( e^{i_{k+1}} \wedge ... \wedge e^{i_n} )$
$= \frac{1}{(n-k)!} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} ( \frac{1}{k!} {\epsilon^{i_{k+1} ... i_n}}_{j_1 ... j_k} e^{j_1} \wedge ... e^{j_k} )$
$= \frac{1}{k! (n-k)!} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} {\epsilon^{i_{k+1} ... i_n}}_{j_1 ... j_k} e^{j_1} \wedge ... e^{j_k}$
Next we have to move the $j_1 ... j_k$ indexes to the front. Moving a single $j_1$ takes $(n-k)$ flips, so moving $k$ of them takes $k (n-k)$ flips:
Equivalently we could move the $i_1 ... i_k$ to the back, and it would require just as many flips. Is this an argument for the Generalized Kronecker delta being defined by either the a front-aligned or back-aligned indexes?
$= (-1)^{k(n-k)} \frac{1}{k! (n-k)!} {\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} {\epsilon_{j_1 ... j_k}}^{i_{k+1} ... i_n} e^{j_1} \wedge ... e^{j_k}$
All duplicates of the Levi-Civita product over $(n-k)$ indexes = $(n-k)!$. This brings us to the generalized Kronecker delta definition:
... using ${\epsilon^{i_1 ... i_k}}_{i_{k+1} ... i_n} {\epsilon_{j_1 ... j_k}}^{i_{k+1} ... i_n} = \sigma \frac{1}{(n-k)!} \delta^{i_1 ... i_k}_{j_1 ... j_k}$
$= (-1)^{k(n-k)} \frac{1}{k!} \sigma \delta^{i_1 ... i_k}_{j_1 ... j_k} e^{j_1} \wedge ... e^{j_k}$
Replace with its definition:
$= (-1)^{k(n-k)} \frac{1}{k!} \cdot k! \sigma \delta^{i_1}_{[j_1} \cdot ... \cdot \delta^{i_k}_{j_k]} e^{j_1} \wedge ... e^{j_k}$
$= \sigma (-1)^{k(n-k)} \delta^{i_1}_{[j_1} \cdot ... \cdot \delta^{i_k}_{j_k]} e^{j_1} \wedge ... e^{j_k}$
Now consider that the wedge is already antisymmetric:
$= \sigma (-1)^{k(n-k)} \delta^{i_1}_{j_1} \cdot ... \cdot \delta^{i_k}_{j_k} e^{j_1} \wedge ... e^{j_k}$
And finally apply the Kronecker deltas:
$= \sigma (-1)^{k(n-k)} e^{i_1} \wedge ... e^{i_k}$
Therefor for any k-form:
$\star \star T = \sigma (-1)^{k (n-k)} T$

Therefore for 0-form 1:
$\star \star 1 = \sigma$
And for n-form $\eta$:
$\star \star \eta = \sigma \eta$

Hodge dual inner product of two p-forms:
$a \cdot b$ ... maybe better denoted $\langle a, b \rangle$ for forms, like some texts do?
I'm going to carry this out with a volume element attached to it:
$\star \langle a, b \rangle$
$= a \wedge \star b$
$= (\frac{1}{k!} a_{i_1 ... i_k} e^{i_1} \wedge ... \wedge e^{i_k}) \wedge \star (\frac{1}{k!} b_{j_1 ... j_k} e^{j_1} \wedge ... \wedge e^{j_k})$
$= \frac{1}{(k!)^2} a_{i_1 ... i_k} b_{j_1 ... j_k} e^{i_1} \wedge ... \wedge e^{i_k} \wedge ( \frac{1}{(n-k)!} {\epsilon^{j_1 ... j_k}}_{i_{k+1} ... i_n} e^{i_{k+1}} \wedge ... \wedge e^{i_n})$
$= \frac{1}{(k!)^2 (n-k)!} a_{i_1 ... i_k} b_{j_1 ... j_k} {\epsilon^{j_1 ... j_k}}_{i_{k+1} ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}$
$= \frac{1}{(k!)^2 (n-k)!} a_{i_1 ... i_k} b_{j_1 ... j_k} {\epsilon^{j_1 ... j_k}}_{i_{k+1} ... i_n} \delta^{i_1}_{k_1} \cdot ... \cdot \delta^{i_n}_{k_n} e^{k_1} \wedge ... \wedge e^{k_n}$
$= \frac{1}{(k!)^2 (n-k)!} a_{i_1 ... i_k} b_{j_1 ... j_k} {\epsilon^{j_1 ... j_k}}_{i_{k+1} ... i_n} \frac{1}{n!} \delta^{i_1 ... i_n}_{k_1 ... k_n} e^{k_1} \wedge ... \wedge e^{k_n}$
$= \frac{1}{(k!)^2 (n-k)!} a_{i_1 ... i_k} b_{j_1 ... j_k} {\epsilon^{j_1 ... j_k}}_{i_{k+1} ... i_n} \frac{1}{n!} \sigma \epsilon^{i_1 ... i_n} \epsilon_{k_1 ... k_n} e^{k_1} \wedge ... \wedge e^{k_n}$
... using $\epsilon_{1...n} = \sqrt{|g|}$ so $\epsilon_{1...n} e^1 \wedge ... \wedge e^n = \sqrt{|g|} e^1 \wedge ... \wedge e^n = \star 1$ so $\epsilon_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n} = n! \star 1$...
$= \frac{1}{(k!)^2 (n-k)!} a_{i_1 ... i_k} b_{j_1 ... j_k} {\epsilon^{j_1 ... j_k}}_{i_{k+1} ... i_n} \sigma \epsilon^{i_1 ... i_n} \star 1$
$= \frac{1}{(k!)^2 (n-k)!} \sigma a_{i_1 ... i_k} b^{j_1 ... j_k} \epsilon_{j_1 ... j_k i_{k+1} ... i_n} \epsilon^{i_1 ... i_n} \star 1$
... using $\epsilon_{j_1 ... j_k i_{k+1} ... i_n} \epsilon^{i_1 ... i_n} = \sigma (n-k)! \delta^{i_1 ... i_k}_{j_1 ... j_k}$...
$= \frac{1}{(k!)^2} a_{i_1 ... i_k} b^{j_1 ... j_k} \delta^{i_1 ... i_k}_{j_1 ... j_k} \star 1$
... using $\delta^{i_1 ... i_k}_{j_1 ... j_k} = k! \delta^{i_1}_{[j_1} \cdot ... \cdot \delta^{i_k}_{j_k]}$...
$= \frac{1}{k!} a_{i_1 ... i_k} b^{j_1 ... j_k} \delta^{i_1}_{[j_1} \cdot ... \cdot \delta^{i_k}_{j_k]} \star 1$
$= \frac{1}{k!} a_{i_1 ... i_k} b^{[i_1 ... i_k]} \star 1$
$= \frac{1}{k!} a_{[i_1 ... i_k]} b^{[i_1 ... i_k]} \star 1$

Using $\sigma \star \star T = T$:
$\langle a, b \rangle$
$= \sigma \star (a \wedge \star b)$
$= \frac{1}{k!} a_{[i_1 ... i_k]} b^{[i_1 ... i_k]} \sigma \star \star 1$
$= \frac{1}{k!} a_{[i_1 ... i_k]} b^{[i_1 ... i_k]}$

So $\star \langle a, b \rangle = \frac{1}{k!} a_{[i_1 ... i_k]} b^{[i_1 ... i_k]}$

Is $\langle a, b \rangle$ of forms via the Hodge-dual definition commutative?
$\langle a, b \rangle$
start with $\langle a, b \rangle = \frac{1}{k!} a_{[i_1 ... i_k]} b^{[i_1 ... i_k]}$
$= \frac{1}{k!} a_{[i_1 ... i_k]} b^{[i_1 ... i_k]}$
...by commutativity of multiplication between a, b, and the metric for raising/lowering...
$= \frac{1}{k!} b_{[i_1 ... i_k]} a^{[i_1 ... i_k]}$
...using $\langle a, b \rangle = \frac{1}{k!} a_{[i_1 ... i_k]} b^{[i_1 ... i_k]}$
$\langle b, a \rangle$
So yes, $\langle a, b \rangle$ is commutative.

What is $\langle \star a, \star b \rangle$ related to $\langle a, b \rangle$?
$\langle \star a, \star b \rangle$
...using $\langle a, b \rangle = \sigma \star (a \wedge \star b)$...
$= \sigma \star (\star a \wedge \star \star b)$
...using $\star \star T = \sigma (-1)^{k(n-k)} T$...
$= \sigma \star (\star a \wedge \sigma (-1)^{k(n-k)} b)$
...factor out scalar...
$= \star ((-1)^{k(n-k)} \star a \wedge b)$
...using $a \wedge b = (-1)^{pq} b \wedge a$ for p-form a and q-form b...
$= \star ((-1)^{k(n-k)} (-1)^{(n-k)k} b \wedge \star a)$
... combine exponents...
$= \star (b \wedge \star a)$
...using $\langle a, b \rangle = \sigma \star (a \wedge \star b)$...
$= \sigma \langle b, a \rangle$
...by commutativity of $\langle a, b \rangle$...
$= \sigma \langle a, b \rangle$

... but what if $a \rightarrow \star a$ ?
Does the index-notation definition factor turn into a $\frac{1}{(n-k)!}$ ?
Does $ \frac{1}{k!} a_{[i_1 ... i_k]} b^{[i_1 ... i_k]} = \frac{1}{(n-k)!} \epsilon^{[i_1 ... i_n]} (\star a)_{[i_{k+1} ... i_n]} \epsilon_{[i_1 ... i_k j_{k+1} ... j_n]} (\star b)^{[j_{k+1} ... j_n]} $ ?
Otherwise something seems wrong...

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Exterior Basis Elements

Here's our volume form, a n-form for n-dimensions:
$\eta = \star 1$
By definition of the Hodge star above:
$\eta = \frac{1}{n!} \epsilon_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}$
Courtesy of the $\sqrt{|g|}$ weighting of $\epsilon_{i_1 ... i_n}$ and of there being $n!$ number of ways that the Levi-Civita and the wedge permutations are summed, this is equal to...
$\eta = \sqrt{|g|} e^1 \wedge ... \wedge e^n$
We can also instead use the definition $e^{\wedge I} = k! e^{[I]}$ to find:
$\eta = \epsilon_{i_1 ... i_n} e^{[i_1} \otimes ... \otimes e^{i_n]}$
$\eta = \epsilon_{[i_1 ... i_n]} e^{i_1} \otimes ... \otimes e^{i_n}$
$\eta = \epsilon_{i_1 ... i_n} e^{i_1} \otimes ... \otimes e^{i_n}$

Notice that $\eta$ is a n-form, so its components are:
$\eta = (\eta)_{i_1 ... i_n} e^{i_1} \otimes ... \otimes e^{i_n}$
So $(\eta)_{i_1 ... i_n} = \epsilon_{i_1 ... i_n}$

Next comes our volume (n-1)-forms:
$\eta_{i_1} = e_{i_1} \lrcorner \eta$
$= \epsilon_{j_1 ... j_n} (e^{j_1} \otimes ... \otimes e^{j_n}) (e_{i_1})$
$= \epsilon_{j_1 ... j_n} e^{j_1} (e_{j_1}) \cdot e^{j_2} \otimes ... \otimes e^{j_n}$
$= \epsilon_{j_1 ... j_n} \delta^{j_1}_{i_1} \cdot e^{j_2} \otimes ... \otimes e^{j_n}$
$= \epsilon_{i_1 j_2 ... j_n} e^{j_2} \otimes ... \otimes e^{j_n}$
$= \epsilon_{i_1 j_2 ... j_n} e^{[j_2} \otimes ... \otimes e^{j_n]}$
$= \frac{1}{(n-1)!} \epsilon_{i_1 j_2 ... j_n} e^{j_2} \wedge ... \wedge e^{j_n}$
$= \epsilon_{i_1 |j_2 ... j_n|} e^{j_2} \wedge ... \wedge e^{j_n}$
So $\eta_i$ is a (n-1)-form that spans all except the i'th dimension.
$= g_{i_1 j_1} {\epsilon^{j_1}}_{j_2 ... j_n} e^{j_2} \otimes ... \otimes e^{j_n}$
$= g_{i_1 j_1} \star e^{j_1}$
... $= \star e_i$ ? ... Depends on which convention you are using, if you will allow this possibly-abuse of index notation or not, whether $e_i$ is distinct object of $g_{ij} e^j$, the first being a vector and the second being the linear combination of a form.

So keep going, to define our (n-k)-forms:
$\eta_I = e_I \lrcorner \eta$
$\eta_{i_1 ... i_k} = (e_{i_1} \otimes ... \otimes e_{i_k}) \lrcorner \eta$
$= \epsilon_{i_1 ... i_k j_{k+1} ... j_n} e^{j_{k+1}} \otimes ... \otimes e^{j_n}$
$= \epsilon_{i_1 ... i_k j_{k+1} ... j_n} e^{[j_{k+1}} \otimes ... \otimes e^{j_n]}$
$= \frac{1}{(n-k)!} \epsilon_{i_1 ... i_k j_{k+1} ... j_n} e^{j_{k+1}} \wedge ... \wedge e^{j_n}$
$= \epsilon_{i_1 ... i_k |j_{k+1} ... j_n|} e^{j_{k+1}} \wedge ... \wedge e^{j_n}$
So $\eta_I$ is the (n-k)-form that spans all except the $i_1 ... i_k$ dimensions.
$= g_{i_1 j_1} \cdot ... \cdot g_{i_k j_k} {\epsilon^{j_1 ... j_k}}_{j_{k+1} ... j_n} e^{j_{k+1}} \otimes ... \otimes e^{j_n}$
... using $\star (e^{j_1} \otimes ... \otimes e^{j_k}) = \frac{1}{k!} {\epsilon^{j_1 ... j_k}}_{j_{k+1} ... j_n} e^{j_{k+1}} \otimes ... \otimes e^{j_n}$...
$= g_{i_1 j_1} \cdot ... \cdot g_{i_k j_k} k! \star (e^{j_1} \otimes ... \otimes e^{j_k})$
$= g_{i_1 j_1} \cdot ... \cdot g_{i_k j_k} \star (e^{j_1} \wedge ... \wedge e^{j_k})$
... $= \star (e_{i_1} \wedge ... \wedge e_{i_k})$ ... based on convention.

Until we get to our 1-form:
$\eta_{i_1 ... i_n} = (e_{i_1} \otimes ... \otimes e_{i_n}) \lrcorner \eta$
$\eta_{i_1 ... i_n} = \epsilon_{i_1 ... i_n}$

So in summary, with basis...
$\eta = (\eta)_{i_1 ... i_n} = \epsilon_{i_1 ... i_n} e^{[i_1} \otimes ... \otimes e^{i_n]}$
$\eta_{i_1} = (\eta_{i_1})_{i_2 ... i_n} = \epsilon_{i_1 ... i_n} e^{[i_2} \otimes ... \otimes e^{i_n]}$
$\eta_{i_1 i_2} = (\eta_{i_1 i_2})_{i_3 ... i_n} = \epsilon_{i_1 ... i_n} e^{[i_3} \otimes ... \otimes e^{i_n]}$
...
$\eta_{i_1 ... i_{n-1}} = (\eta_{i_1 ... i_{n-1}})_{i_n} = \epsilon_{i_1 ... i_n} e^{i_n}$
$\eta_{i_1 ... i_n} = \eta_{i_1 ... i_n}$

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Pseudo-Vectors

Often angular-velocity, angular-momentum, torque, etc are called "pseudo-vectors". This is because, like a vector, they represent components in 3D space, but unlike a tensor they are not preserved with regards to change-of-basis.
Here's why.

Is the change-of-basis of the dual of a tensor equal to the dual of the change-of-basis of the tensor?

Change-of-basis of the dual of a tensor:
$|\star \omega^{ij}|$
... change basis first ...
$= |{e_a}^i {e_b}^j \star \omega^{ab}|$
... then perform dual ...
$= |{e_a}^i {e_b}^j {\epsilon^{ab}}_c \omega^c|$
$= |{e_a}^i {e_b}^j g^{au} g^{bv} \epsilon_{uvc} \omega^c|$
$= |{e_a}^i {e_b}^j g^{au} g^{bv} \sqrt{|g|} \bar\epsilon_{uvc} \omega^c|$
$= \sqrt{|g|} \cdot \sqrt{|g|} \cdot \frac{1}{|g|} \cdot \frac{1}{|g|} \cdot \sqrt{|g|} \cdot |\pm 1| \cdot |\vec\omega|$
$= \frac{1}{\sqrt{|g|}} \cdot |\vec\omega|$

Dual of change-of-basis of a tensor:
$|\star \omega^{ij}|$
... perform dual first ...
$= |{\epsilon^{ij}}_k \omega^k|$
... then change basis ...
$= |{\epsilon^{ij}}_k {e_c}^k \omega^c|$
$= |\eta^{im} \eta^{jn} \epsilon_{mnk} {e_c}^k \omega^c|$
$= |\eta^{im} \eta^{jn} \sqrt{|\eta|} \bar\epsilon_{mnk} {e_c}^k \omega^c|$
$= \frac{1}{|\eta|} \cdot \frac{1}{|\eta|} \cdot \sqrt{|\eta|} \cdot |\pm 1| \cdot \sqrt{|g|} \cdot |\vec\omega|$
$= \frac{1}{|\eta|^{3/2}} \cdot \sqrt{|g|} \cdot |\vec\omega|$
... using $|\eta| = 1$...
$= \sqrt{|g|} \cdot |\vec\omega|$

So the lesson here is: if you need to change-of-basis some kind of dual-tensor of a physical property, like a rotation, i.e. angular momentum, torque, etc, then use your original tensors or exterior products, not their dual angle-axis vectors.

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What if if you have a sum-of-p-forms on a n-form wedge space?

For $w = w^0 + {w^1}_i e^i + \frac{1}{2} {w^2}_{i_1 i_2} e^{i_1} \wedge e^{i_2} + ... + \frac{1}{n!} {w^n}_{i_1 ... i_n} e^{i_1} \wedge ... \wedge e^{i_n}$

... and you want to eliminate all except the p'th form?

$w \wedge e^{j_{p+1}} \wedge ... \wedge e^{j_n}$
... will zero everything of form > p-form ...
$ = (w^0 + {w^1}_i e^i + \frac{1}{2} {w^2}_{i_1 i_2} e^{i_1} \wedge e^{i_2} + ... + \frac{1}{p!} {w^p}_{i_1 ... i_p} e^{i_1} \wedge ... \wedge e^{i_p}) \wedge e^{j_{p+1}} \wedge ... \wedge e^{j_n}$
... next dual everything, so that the p-form maps to the (n-p)-form, and it becomes the smallest form instead of the largest ...
... then wedge with $e^{k_1} \wedge ... \wedge e^{k_{n-p-1}}$, so that everything else goes away, and you are left with ...
$= \frac{1}{p!} {w^p}_{i_1 ... i_p} \star (e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_{p+1}} \wedge ... \wedge e^{j_n}) \wedge e^{k_1} \wedge ... \wedge e^{k_{n-p-1}}$
$= \frac{1}{p!} {w^p}_{i_1 ... i_p} \epsilon^{i_1 ... i_p j_{p+1} ... j_n}) \wedge e^{k_1} \wedge ... \wedge e^{k_{n-p-1}}$
... now wedge with $e^{k_{n-p}} \wedge ... \wedge e^{k_n}$ ... and then dual times signature ... and we have isolated this term.

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Cross product in $\mathbb{R}^3$:

Let $dx, dy, dz$ form our one-form basis of $\mathbb{R}^3$

Let $a = a_1 dx^1 + a_2 dx^2 + a_3 dx^3$ and $b = b_1 dx^1 + b_2 dx^2 + b_3 dx^3$
$a \wedge b$
$= (a_1 dx^1 + a_2 dx^2 + a_3 dx^3) \wedge (b_1 dx^1 + b_2 dx^2 + b_3 dx^3)$
$= a_1 dx^1 \wedge (b_1 dx^1 + b_2 dx^2 + b_3 dx^3) + a_2 dx^2 \wedge (b_1 dx^1 + b_2 dx^2 + b_3 dx^3) + a_3 dx^3 \wedge (b_1 dx^1 + b_2 dx^2 + b_3 dx^3) $
$= a_1 dx^1 \wedge b_1 dx^1 + a_1 dx^1 \wedge b_2 dx^2 + a_1 dx^1 \wedge b_3 dx^3 + a_2 dx^2 \wedge b_1 dx^1 + a_2 dx^2 \wedge b_2 dx^2 + a_2 dx^2 \wedge b_3 dx^3 + a_3 dx^3 \wedge b_1 dx^1 + a_3 dx^3 \wedge b_2 dx^2 + a_3 dx^3 \wedge b_3 dx^3 $
using $dx^i \wedge dx^i = 0$ and $dx^i \wedge dx^j = -dx^j \wedge dx^i$
$= (a_2 b_3 - a_3 b_2) dx^2 \wedge dx^3 + (a_3 b_1 - a_1 b_3) dx^3 \wedge dx^1 + (a_1 b_2 - a_2 b_1) dx^1 \wedge dx^2 $

Ta da, the cross product emerges. And if we want it in the original basis?

$\star (a \wedge b)$
$= \star ( (a_2 b_3 - a_3 b_2) dx^2 \wedge dx^3 + (a_3 b_1 - a_1 b_3) dx^3 \wedge dx^1 + (a_1 b_2 - a_2 b_1) dx^1 \wedge dx^2 )$
$= (a_2 b_3 - a_3 b_2) \star (dx^2 \wedge dx^3) + (a_3 b_1 - a_1 b_3) \star (dx^3 \wedge dx^1) + (a_1 b_2 - a_2 b_1) \star (dx^1 \wedge dx^2) $
$= (a_2 b_3 - a_3 b_2) dx^1 + (a_3 b_1 - a_1 b_3) dx^2 + (a_1 b_2 - a_2 b_1) dx^3 $

Proof that the famous a cross b dot c equation equals the volume of a parallelepiped in 3D:
Volume of a parallelepiped in 3D:
$(a \times b) \cdot c)$
using $a \times b = \star (a \wedge b)$
$= (\star (a \wedge b)) \cdot c$
$= c \cdot (\star (a \wedge b))$
using $a \cdot b = \star (a \wedge \star b)$
$= \star (c \wedge \star (\star (a \wedge b)))$
$= \star (c \wedge a \wedge b)$
$= \star (a \wedge b \wedge c)$


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