Covariant derivative of Levi-civita symbol

To clarify a little, the Levi-Civita symbol $epsilon_{abc}$ is a number, the one you wrote down, and not a tensor. It is a generalisation of the Kronecker delta. You can use it to build a tensor, the Levi-Civita tensor $varepsilon$, whose components you wrote down. So okay, maybe it's a pseudo-tensor. It is the canonical volume form on the Riemannian manifold.

To even talk about covariant differentiation, there must be an implicit connexion and hence Christoffel symbols. Well, you have assumed a metric $g_{ij}$ so okay, there is an important theorem that says that even though there are many different possible connexions, there is only one connexion on a Riemannian manifold that is

(a) has no torsion, so it is symmetric, and

(b) it is consistent with the metric in that the covariant derivative (using that connexion) of the metric tensor is zero.

Another basic theorem of Riemannian geometry says that there is, at any one point $p$ that you decide on (and then do not get to change), there is a coordinate system such that all the Christoffel symbols vanish (but only at that point, they do vary in a neighbourhood of p, no matter how small) and all the first derivatives of the metric tensor's components $g_{ij}$ vanish at $p$.

Now we are ready to answer your question, and since the question is covariant, independent of the coordinates, our answer will be generally true for any coordinate system even though we are doing our calculation in a very special coordinate system.

The volume form is unique so even in these coordinates, it is $sqrt{-g}$dx ^ dy ^dz. And since the Christoffel symbols are all zero at $p$, the covariant derivative in any direction, e.g. the three coordinate directions, is given by the usual partial derivative of these coordinates.

With this set up, the ideas that other posters have attempted to communicate now work. Consider any directional derivative of $sqrt{-g}$dx ^ dy ^dz, e.g., $frac partial {partial x} (sqrt{-g}dx wedge dy wedge dz)$. By the Leibniz rule, for covariant derivatives, it is $(frac partial {partial x} sqrt{-g}) dx wedge dy wedge dz + sqrt{-g}frac partial {partial x}(dx wedge dy wedge dz)$.
As I remarked, these derivatives can be either the $x$-component of the covariant derivative of the tensor or the ordinary partial derivative with respect to x of the component of the tensor, because the Christoffel symbols are zero.

But the partial of $g$ is zero since we assumed all first derivatives of the $g_{ij}$ in our coordinate system vanished. (It is also true that the covariant derivative of the metric tensor, and hence any function of it, also vanishes, in any coordinate system, but this is why.) So the first summand vanishes.

The second summand vanishes since all its components vanish: the coefficients of dx ^ dy ^dz are constant, so their partials vanish. This proves your formula.