One way to see this is to consider the fact that the vector space of rank (3,3) completely antisymmetric tensors ($ \Lambda_3^3(R^3) $) has dimension one (it's just a linear algebra exercise). Then define the tensor:
$$
M_{ijk}^{lmn} = \delta_i^{[l} \, \delta_j^m \, \delta_k^{n]} = \frac{1}{3!} \sum_{\sigma \in S_3} sgn(\sigma) \, \delta_i^{\sigma(l)} \, \delta_j^{\sigma(m)} \, \delta_k^{\sigma(n)}
$$
where we are summing over all the permutations of three numbers $\sigma$, and $sgn(\sigma)$ denotes the sign of the permutation. It's worthy to note that
$$
M_{ijk}^{lmn}=\frac{1}{3!}\begin{vmatrix} \delta_i^l & \delta_i^m & \delta_i^n\\
\delta_j^l & \delta_j^m & \delta_j^n \\ \delta_k^l & \delta_k^m & \delta_k^n \end{vmatrix}
$$
by the Leibniz formula of the determinant (http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants). So we have $M \in \Lambda_3^3(R^3) $. Since $M \neq 0$, $M$ is a basis for the space $ \Lambda_3^3(R^3) $. Now consider the tensor
$$
\epsilon_{ijk} \, \epsilon^{lmn} = B_{ijk}^{lmn}
$$
Since $B \in \Lambda_3^3(R^3) $ and $M$ is a basis, there exists a constant $k$ such that
$$
B_{ijk}^{lmn} = k \, M_{ijk}^{lmn} \implies \epsilon_{ijk} \, \epsilon^{lmn} = k \, \delta_i^{[l} \, \delta_j^m \, \delta_k^{n]}
$$
Now to determine $k$ contract $\epsilon_{lmn}$ on both sides and use the fact that $\epsilon_{lmn} \, \epsilon^{lmn} =3!$ (since you sum $3!$ terms equal to one)
$$
\epsilon_{ijk} \, \epsilon^{lmn} \, \epsilon_{lmn} = 3! \, \epsilon_{ijk} = k \, \delta_i^{[l} \, \delta_j^m \, \delta_k^{n]} \epsilon_{lmn} = k \, \delta_i^{l} \, \delta_j^m \, \delta_k^{n} \, \epsilon_{[lmn]} = k \, \delta_i^{l} \, \delta_j^m \, \delta_k^{n} \, \epsilon_{lmn} = k \, \epsilon_{ijk}
$$
So we finally get $k=3!$ and
$$
\epsilon_{ijk} \, \epsilon^{lmn}=\begin{vmatrix} \delta_i^l & \delta_i^m & \delta_i^n\\
\delta_j^l & \delta_j^m & \delta_j^n \\ \delta_k^l & \delta_k^m & \delta_k^n \end{vmatrix}
$$
And you can get the identity you want by contracting. The same argument could be used in any dimension to show that
$$
\epsilon_{i_1,\dots,i_n} \, \epsilon^{j_1,\dots,j_n} = \begin{vmatrix} \delta_{i_1}^{j_1} & \dots & \delta_{i_1}^{j_n}\\
\vdots & & \vdots \\ \delta_{i_n}^{j_1} & \dots & \delta_{i_n}^{j_n} \end{vmatrix}
$$
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 (pseudo-)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 (pseudo-)Riemannian manifold that:
(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), 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 generally be 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 $\sqrt{-g}\wedge dy \wedge dz$.
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 $\sqrt{-g} dx \wedge dy \wedge dz$, 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 $dx \wedge dy \wedge dz$ are constant, so their partials vanish. This proves your formula.
Best Answer
This thread on physicsforums elaborates a bit on the difference between Levi-Civita symbols and tensors. Based on that, I conclude...
1) Your index notation formula for the magnetic field should use the Levi-Civita tensor, then. The "symbol" is a convenient thing, but this expression must be written with tensors.
2) Carroll likely made a mistake and meant to talk about the Levi-Civita tensor's transformation properties.
3) Any expression where the same index appears on the bottom twice (or the top twice) is just laziness on the part of the author. It's a common laziness, especially in contexts where one doesn't discriminate between covariant and contravariant components, however.
Actually, I'm not sure what your question is here.
4) Again, I would say that this expression should use the tensor, not the symbol.
5) I see no reason why you wouldn't.
Just some general remarks on the Levi-Civita tensor/symbol and what they represent: a flat space has a unique "volume form" or "pseudoscalar". There is a unit volume element that all other volumes are scalar multiples of. This volume has an orientation (think of a volume spanned by vectors according to the right-hand rule vs. a left-hand rule).
The Levi-Civita tensor and symbol are related to this notion. The tensor represents the components of the unit volume element with respect to volume elements built by combinations of basis vectors.
The volume element can be used to perform duality operations. This is the foundation of the Hodge star operator. Using the $N$-dimensional Levi-Civita tensor on a tensor object with $k$ free indices yields a new object with $N-k$ free indices. This can be described geometrically, too. In 3d, scalars would go to volumes, vectors to planes, planes to vectors, and volumes to scalars under this duality operation. It is the explicit mapping of planes to vectors that is so often performed with duality--it allows us to cheat in 3d by dealing with only scalars and vectors. Planes and volumes can then be mapped back to vectors and scalars by duality. This is exactly what is done with the magnetic field and angular momentum. You should see clearly that both of these vectors, if not for the use of the Levi-Civita tensor, would be expressions with 2 free indices, and antisymmetric on those indices. These objects are called bivectors.
The Levi-Civita tensor and symbol are often used in physics and math to treat expressions through duality rather than directly--even when, in my opinion, this obscures the real physics of the problem or covers up for a shortcoming of the notation. Just the other day around here we had a question about building up 4-volumes from a single plane. Geometrically, this is obvious--you can't build a 4-volume from a single plane. But in index notation, it was cumbersome at best, involving finding the dual plane through use of the Levi-Civita tensor and taking traces.
Overall, the Levi-Civita tensor and its many indices can be difficult to work with, especially in arbitrary coordinate systems. I once heard a professor bemoan that all the identities another professor had taught students with Levi-Civita had only used the symbol--i.e. the tensor in cartesian coordinates--and so they weren't valid in arbitrary coordinate systems. The solution suggested was to teach students about tensor densities, which was met with skepticism at best, since there were only three professors in the whole department that, in the other professor's view, either cared about or even knew about tensor densities. I think part of this view is why the Levi-Civita symbol is often used instead; it's just easier to prove some things in cartesian coordinates, even if the resulting expression is not really correct (not really a tensor because the metric determinant has been ignored, etc.).