Recall that the Lorentz group $O(1,3)$ consists of all $4\times 4$ matrices $\Lambda$ satisfying $\Lambda^T\eta\Lambda=\eta$. This gives, in particular, $\det(\Lambda)=\pm1$. The subgroup $SO(1,3)$ consists of all transformations with determinant $+1$.
$\vec{E}\cdot\vec{B}$ is invariant under $SO(1,3)$ but not under all of $O(1,3)$ - it's a pseudoscalar. This is because it's a product of the vector $\vec{E}$ and the pseudovector $\vec{B}$. Here's an easy way of seeing that $\vec{B}$ transforms as a pseudovector:
Consider the Lorentz force law $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$. Now apply a reflection: \begin{equation}R:(t,\vec{x})\mapsto(t,-\vec{x}).\end{equation} We have: \begin{equation}\vec{F}\mapsto-\vec{F} \qquad \text{and} \qquad\vec{E}\mapsto-\vec{E},\end{equation} so that $\vec{v}\times\vec{B}\mapsto-\vec{v}\times\vec{B}$.
But the velocity transforms as $\vec{v}\mapsto-\vec{v}$ so we have the transformation of $\vec{B}$ under $R$:\begin{equation}\vec{B}\mapsto\vec{B}=(-1)(-\vec{B})=\det(R)(R\cdot\vec{B}).\end{equation}
You can use Young tableaux/diagrams and the permutation group to figure out the symmetries of the general rank-3 tensor. The spaces correspond to the partitions of the rank:
3=3:
One 20 dimensional total symmetric subspace.
3=2+1:
Two 20 dimensional mixed symmetry subspaces.
3=1+1+1:
One 4 dimensional totally antisymmetric subspace:
$$ A_{\alpha\beta\gamma} = \frac 1 6 [T_{\alpha\beta\gamma} + T_{\beta\gamma\alpha} + T_{\gamma\alpha\beta} - T_{\gamma\beta\alpha} - T_{\beta\alpha\gamma} - T_{\alpha\gamma\beta}] $$
That is the only antisymmetric thing you can make according to Schurl-Weyl theory.
To find the dimensions, I used the Hook Length Formula (summing of the boxes $x$ in a diagram $Y(\lambda)$) for the Young diagram corresponding to the integer partition:
$$ {\rm dim}\pi_{\lambda} = \frac {n!}{\prod_{x\in Y}{\rm hook}(x)}$$
If you consider 3 dimensions ($n=3$), you get ${\rm dim} = 1$, that is the standard Levi-Civita symbol $\epsilon_{ijk}$.
If you set $n=4$, the result is ${\rm dim} = 4$.
That means $A_{\alpha\beta\gamma}$ transforms like a 4-vector.
So, the only antisymmetric part of a rank-3 tensor in Minkowski space rotates like a 4-vector, which means it is not invariant and is not a candidate to be Levi-Civita like.
Meanwhile, the dimensions of the 3 other irreducible spaces are all 20--which are certainly not scalars, and thus not candidates to be Levi-Civita like.
Note that if you consider rank-4 tensors, the partitions are as follows:
4=4:
35 dimensional and symmetric.
4=3+1:
Three 45-dimensional mixed symmetry spaces.
4=2+2:
Two 20-dimensional mixed symmetry spaces.
4=2+1+1:
Three 15-dimensional mixed symmetry spaces.
4=1+1+1+1:
One total antisymmetric 1 dimensional space, which is proportional to the Levi-Civita symbol $\epsilon_{\mu\nu\sigma\lambda}$.
In summary, the answer is "No", and the reason why has to do with the representations of the symmetric group on 3-letters. You partition the rank=3, use the Robinson-Schensted correspondence to associate that partition with irreducible representations of the permutation group. (The Young Diagrams make this step a snap). Then, Schur-Weyl duality associates those with irreducible subspaces of and rank-N tensor (signed permutations of indices). Finally, the Hook Length formula tells you the dimensions of those subspaces.
The Levi-Civita symbol needs to be invariant (e.g., dimension 1, like a scalar) and it need to be totally antisymmetric in all indices--and that simply did not exist for rank 3 in 4 dimensions.
Best Answer
If I understand it correctly, you want to prove that $ \epsilon^{\mu\nu\sigma\rho} \partial_\rho F_{\nu\sigma} = 0 $ for a general anti-symmetric $F_{\nu\sigma}$.
From Bianchi Identity we have : $$ \epsilon^{\mu\nu\sigma\rho} (\partial_\rho F_{\nu \sigma} + \partial_\nu F_{\sigma \rho} + \partial_\sigma F_{\rho \nu}) = 0 \\ \epsilon^{\mu\nu\sigma\rho}\partial_\rho F_{\nu \sigma} + \epsilon^{\mu\nu\sigma\rho}\partial_\nu F_{\sigma \rho} + \epsilon^{\mu\nu\sigma\rho}\partial_\sigma F_{\rho \nu} = 0 \\ $$
Now realise that in the all the terms the indices are contracted (hence can be replaced by other indices) and write all the terms such that the indices on $\partial$ and F are same in each term so that the $\epsilon$ indices get modified :
$$ \epsilon^{\mu\nu\sigma\rho}\partial_\rho F_{\nu \sigma} + \epsilon^{\mu\rho\nu\sigma}\partial_\rho F_{\nu \sigma} + \epsilon^{\mu\sigma\rho\nu}\partial_\rho F_{\nu \sigma} = 0 \\ (\epsilon^{\mu\nu\sigma\rho}+\epsilon^{\mu\rho\nu\sigma}+\epsilon^{\mu\sigma\rho\nu}) \partial_\rho F_{\nu \sigma} = 0 $$
Now, use the fact that $\epsilon$ is fully anti-symmetric in its indices so that :
$$ \epsilon^{\mu\nu\sigma\rho} = \epsilon^{\mu\rho\nu\sigma} =\epsilon^{\mu\sigma\rho\nu} $$
So, we get :
$$ \epsilon^{\mu\sigma\rho\nu}\partial_\rho F_{\nu \sigma} =0 $$
Also, again exchanging indices we get :
$$ \epsilon^{\mu\nu\sigma\rho}\partial_\rho F_{\nu \sigma} =0 $$