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.).
For a proof that the only ${\rm SO}(N)$ invariant tensors are products of $\delta_{ab}$'s and Levi-Civita symbols see M. Spivak, A Comprehensive Introduction to Differential Geometry (second edition) Vol. V, pp. 466-481. The number of pages required for the argument shows that it is not trivial.
I've just looked up the third edition of Spivak vol V. What is needed is theorem 35 on page 327. This is the section entitled "A Smattering of Classical Invariant Theory." He writes in terms of scalar invariants, but of course an invariant tensor becomes a scalar invariant when contracted with enough vectors, and any such scalar invariant arises from an invariant tensor.
Best Answer
I suspect that what you are really after is that $\epsilon$ is invariant under Lorentz transformations between inertial observes. In fact, we have a well known formula for the determinat of an operator: $$ \epsilon_{i_1...i_n}A^{i_1}_{\ \ j_1}...A^{i_n}_{\ \ j_n} = \mathrm{det} A \epsilon_{j_1...j_n}. $$ All Lorentz transformations have $\mathrm{det} A=\pm 1$. If you restrict attention to proper ortochronous transformations (basically exclude various reflections) you get that $\epsilon$ is invariant. We say that it is pseudoscalar with respect to Lorentz group. On the other hand, as pointed by others in this thread, it is not a tensor with respect to general curvilinear transformations.