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.
Following the method in L&L vol. 2 ch. 1, the left-hand side
$$\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma}$$
is a product of a pseudo-tensor, which is invariant under Lorentz transformations up to a factor of a determinant, and a pseudo-tensor which transforms under the inverse determinant.
This tells us that the right-hand side has to be an invariant tensor and so must be constructed from Kronecker delta's.
The right hand side must also be anti-symmetric in $\mu,\nu$ and anti-symmetric in $\rho,\sigma$ so that on general grounds
$$
\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma} = - A (\delta^{\mu}_{\rho} \delta^{\nu}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\nu}_{\rho})$$
must hold, where the minus sign is due to the Minkowski metric, and the factor $A=2$ is fixed by the requirement that $\varepsilon^{\mu \nu \rho \sigma} \varepsilon_{\mu \nu \rho \sigma} = - 4!$ holds, so setting $ - A (\delta^{\mu}_{\mu} \delta^{\nu}_{\nu} - \delta^{\mu}_{\nu} \delta^{\nu}_{\mu}) = - 4!$ gives $A(16 - 4) = 24$ or $A = 2$.
More generally, the same thinking tells us that the product $\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma}$ must be an anti-symmetric combination of Kronecker delta's up to an overall normalization factor, where the normalization factor must respect the fact that $\varepsilon^{\mu \nu \rho \sigma} \varepsilon_{\mu \nu \rho \sigma} = - 4!$ should hold, but $\delta^{\mu \nu \rho \sigma}_{\alpha \beta \gamma \delta}$ is an anti-symmetric combination of Kronecker delta's such that $\delta^{\mu \nu \rho \sigma}_{\mu \nu \rho \sigma} = 4!$, and so
$$
\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma} = - \delta^{\mu \nu \rho \sigma}_{\alpha \beta \rho \sigma}$$
must hold. Thus we can work out contractions like $\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma} = - \delta^{\alpha \beta \mu \nu}_{\alpha \beta \rho \sigma}$ directly
\begin{align*}
\varepsilon^{\alpha \beta \mu \nu} \varepsilon_{\alpha \beta \rho \sigma} &= - \delta^{\alpha \beta \mu \nu}_{\alpha \beta \rho \sigma} \\
&= - (\delta^{\alpha}_{\alpha} \delta^{\beta \mu \nu}_{\beta \rho \sigma} - \delta^{\alpha}_{\beta} \delta^{\beta \mu \nu}_{\alpha \rho \sigma} + \delta^{\alpha}_{\rho} \delta^{\beta \mu \nu}_{\alpha \beta \sigma} - \delta^{\alpha}_{\sigma}\delta^{\beta \mu \nu}_{\alpha \beta \rho}) \\
&= - (4\delta^{\beta \mu \nu}_{\beta \rho \sigma} - \delta^{\alpha \mu \nu}_{\alpha \rho \sigma} + \delta^{\beta \mu \nu}_{\rho \beta \sigma} - \delta^{\beta \mu \nu}_{\sigma \beta \rho}) \\
&= - [4(\delta^{\beta}_{\beta} \delta^{\mu \nu}_{\rho \sigma} - \delta^{\beta}_{\rho} \delta^{\mu \nu}_{\beta \sigma} + \delta^{\beta}_{\sigma}\delta^{\mu \nu}_{\beta \rho}) - ( \delta^{\alpha }_{\alpha} \delta^{\mu \nu}_{\rho \sigma} - \delta^{\alpha}_{\rho} \delta^{\mu \nu}_{\alpha \sigma} + \delta^{\alpha}_{\sigma} \delta^{\mu \nu}_{\alpha \rho}) - \delta^{\beta \mu \nu}_{\beta \rho \sigma} + \delta^{\beta \mu \nu}_{\beta \sigma \rho}] \\
&= - [4(4\delta^{\mu \nu}_{\rho \sigma} - \delta^{\mu \nu}_{\rho \sigma} + \delta^{\mu \nu}_{\sigma \rho}) - ( 4 \delta^{\mu \nu}_{\rho \sigma} - \delta^{\mu \nu}_{\rho \sigma} + \delta^{\mu \nu}_{\sigma \rho}) - \delta^{\beta \mu \nu}_{\beta \rho \sigma} + \delta^{\beta \mu \nu}_{\beta \sigma \rho}] \\
&= - [4(2\delta^{\mu \nu}_{\rho \sigma} ) - 2 \delta^{\mu \nu}_{\rho \sigma} - 2 \delta^{\mu \nu}_{\rho \sigma} + 2 \delta^{\mu \nu}_{\sigma \rho}] \\
&= - [8 \delta^{\mu \nu}_{\rho \sigma} - 6 \delta^{\mu \nu}_{\rho \sigma} ] \\
&= - 2 \delta^{\mu \nu}_{\rho \sigma} \\
&= - 2 (\delta^{\mu}_{\rho} \delta^{\nu}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\nu}_{\rho}).
\end{align*}
Similarly, in $d$ dimensions we have
$$\varepsilon^{\mu_1 .. \mu_d} \varepsilon_{\mu_1 .. \mu_d} = - d!$$
From this one immediately sees that
$$
\varepsilon^{\mu_1 .. \mu_d} \varepsilon_{\nu_1 .. \mu_d} = - \delta^{\mu_1 .. \mu_d}_{\nu_1 .. \nu_d}
$$
and so contractions obey identities like
$$\varepsilon^{\mu_1 \ldots \mu_r \mu_{r+1} .. \mu_d} \varepsilon_{\mu_1 .. \mu_r \nu_{r+1} \ldots \nu_d} = - A \delta^{\mu_{r+1} .. \mu_d}_{\nu_{r+1} .. \nu_d}$$
must hold in $d$ dimensions, where $A$ can be fixed by expanding
$$\varepsilon^{\mu_1 \ldots \mu_r \mu_{r+1} .. \mu_d} \varepsilon_{\mu_1 .. \mu_r \nu_{r+1} \ldots \nu_d} = - \delta^{\mu_1 .. \mu_r \mu_{r+1} .. \mu_d}_{\mu_1 .. \mu_r \nu_{r+1} .. \nu_d}$$
one step at a time as in the example above, and obviously $A = r!$ should hold so that $A = - d!$ when $r = d$, as can be proven by induction.
Best Answer
The first equation only holds for Minkowski spacetime, something that explicitly appears in Weinberg's "Gravitation and Cosmology" book (which the linked site references). In that case $\sqrt{-g} = 1$.
The minus sign comes from the fact that, on a Lorentzian manifold, contravariant and covariant tensors may exhibit a sign difference (since the metric is no longer positive-definite). In our case, a Minkowski spacetime will either have 1 or 3 coordinate directions with a negative sign, depending on the convention; $(- + + +)$ and $(+ - - -)$ respectively. Hence after raising the indices:
\begin{equation} \varepsilon ^{\mu \nu \rho \sigma} = g^{\mu \alpha} g^{\nu \beta} g^{\rho \gamma} g^{\sigma \delta} \, \varepsilon _{\alpha \beta \gamma \delta} \Rightarrow \, \varepsilon ^{\alpha \beta \gamma \delta} = -\varepsilon _{\alpha \beta \gamma \delta} \end{equation}
the contravariant Levi-Civita tensor picks up a minus sign. Bear in mind that the notation $[\alpha ,\, \beta \, , \ldots , \, \mu ]$ implies that the indices are lower/raised appropriately in each of the two cases. This means that between your equations (2) and (3), it is implied that there is a relative sign difference. It is not shown because I assume the text doesn't derive (3) from (2), so the indices are not necessarily the same (the identical naming is incidental).
To derive that relative sign difference, let's examine a more general case. For a general spacetime, notice how the product $g^{\mu \alpha} g^{\nu \beta} g^{\rho \gamma} g^{\sigma \delta}$ (assuming that the indices $\mu$, $\nu$, $\rho$ and $\sigma$ as such so that the contravariant Levi-Civita tensor in not 0) along with the implicit sum due to Einstein notation will produce the inverse of the determinant of the metric. This is especially easy to see in the case of a diagonal metric. Therefore one may write:
\begin{equation} \begin{array}{lr} \varepsilon _{\alpha \beta \gamma \delta} = \sqrt{-g} \, [\alpha , \, \beta ,\, \gamma ,\, \delta ] \\ \varepsilon ^{\alpha \beta \gamma \delta} = g^{-1} \, \varepsilon _{\alpha \beta \gamma \delta} \end{array} \Bigg \} \Rightarrow \, \varepsilon ^{\alpha \beta \gamma \delta} = -\frac{[\alpha , \, \beta ,\, \gamma ,\, \delta ]}{\sqrt{-g}} \end{equation}
Had we started backwards and defined:
\begin{equation} \varepsilon ^{\alpha \beta \gamma \delta} = \frac{[\alpha , \, \beta ,\, \gamma ,\, \delta ]}{\sqrt{-g}} \end{equation}
then it would be the covariant tensor that would pick up a minus sign in this convention.