Is the vector cross product only defined for 3D?

Yes, you are correct. You can generalize the cross product to $n$ dimensions by saying it is an operation which takes in $n-1$ vectors and produces a vector that is perpendicular to each one. This can be easily defined using the exterior algebra and Hodge star operator http://en.wikipedia.org/wiki/Hodge_dual: the cross product of $v_1,\ldots,v_{n-1}$ is then just $*(v_1 \wedge v_2 \cdots \wedge v_{n-1}$).

Then the magnitude of the cross product of n-1 vectors is the volume of the higher-dimensional parallelogram that they determine. Specifying the magnitude and being orthogonal to each of the vectors narrows the possiblity to two choices-- an orientation picks out one of these.

The answer to this problem is sadly not very well-known, since it depends on what you really want as "cross product".

1st solution. r-ary operation in any dimension with certain axiomns. Suppose an r-ary operation on certain d-dimensional space V. Then, a r-fold d-dimensional "cross product" multilinear operation exists:

$$ (C_1\times C_2\times \ldots\times C_r): V^{dr}=\underbrace{V^d\times \cdots \times V^d}_{r}\longrightarrow V^d$$

such as $$\forall i=1,2,...,r$$ we have that

$$ (C_1\times C_2\times \ldots\times C_r)\cdot C_i=0$$

$$ (C_1\times C_2\times \ldots\times C_r)\cdot (C_1\times C_2\times \ldots\times C_r)=\det (C_i\cdot C_j)$$

Eckmann (1943) and Whitehead (1963) solved this problem in the continuous case over real euclidean spaces, while Brown and Gray (1967) solved the multilinear case. Moreover, the solution that I am going to provide is valid in any field with characteristic different of 2 and with $1\leq r\leq d$. The theorem (due to Eckmann, Whitehead, and Brown-Gray) says that the "generalized cross product" (including the 3d case) exists when:

A) $d$ is even, $r = 1$. A cross product exists in every even dimension with one single factor. This can be thought some kind of "Wick rotation" if you are aware of this concept in every even dimensions! This cross product with a single factor is a bit non-trivial but easy to understand.

B) $d$ is arbitrary, $r = d − 1$. A cross product exists in arbitrary dimension d and (d-1) factors. It is also said that an arbitrary (d-1)-fold cross product exists in any dimension. Just take the determinant of those (d-1) vectors with the versors $(e_1,...,e_r)$!

C) $d = 3, 7, r = 2$. A 2-fold cross vector exists in dimension 3 and 7. Therefore, the "bilinear" cross product can only exists with two factors in 3D and 7D. The 3D cross product is well known, the 7D cross product can be found (both in coordinate and free coordinate versions) in wikipedia.

D) $d = 8, r = 3$. A 3-fold cross product exists in eight dimensions. There, there is a non-trivial 3-fold cross in 8D, i.e., you can build a non-trivial cross product with 3 vectors in 8 dimensions. I haven't seen a coordinate expression for this but I believe someone did it (I could write a post about it, though, in my blog, in the near future).

This happens in euclidean signature, I suppose there are some variants in pseudo-euclidean metrics (and perhaps some non-trivial subcases; I have heard about a non-trivial 3-fold cross product in 4D but I can not find a reference). Moreover, you can find a similar conclusion in the book Clifford algebras and Spinors by P. Lounesto. Geometric algebra is very useful when handling with this vector stuff since vectors are just a particular grade of a polyvector/cliffor/blade...

2nd solution. Cross product can be seen as the dual of the exterior product via $ia\times b=a\wedge b$ or $\star (a\wedge b)=a\times b$. Therefore, the wedge product (exterior product, a bivector) is much more fundamental since it can be defined in ANY spacetime dimension. You can of course identify bivectors with antisymmetric matrices too, but that is only a realization of the bivector. Indeed, bivector defines rotations in a give plane and this is much more useful than thinking in terms of a vector. Bivectors are the generators of rotations in N-dimensional spaces (even if you consider multivectors or polyvectors fields). Thus, the second solution is to consider the exterior product as the true generalization (with two factors!) of the cross product in any spacetime dimension.

3rd solution. Use k-forms (k-vectors) and give up the 2-ary condition., assuming a metric can be defined. If you keep wanting a VECTOR, or 1-blade, then using the Hodge star operator: $$V=\star(V_1\wedge\cdots \wedge V_{N-1})$$ This produces a 1-form (1-vector) from an N-1-form, N-1-vector. Indeed, if you have a non-factorizable nonsimple k-vector on N-dimensional space, the star operator, with a single term!, produces a (N-k)-form or (N-k)-vector in general, as said in the references of other users.

Well, it depends on what you mean by "the vector cross product." There is a generalization to $n$ dimensions which takes $n-1$ vectors as input and returns what can be thought of as a vector orthogonal to all of them. It generalizes to an operation taking $k$ vectors as input where $k \le n$, but then the output is not something like a vector but something more complicated. See wedge product.

There is a more specific generalization to $7$ dimensions coming from multiplication in the octonions in the same way that the cross product can be thought of as coming from multiplication in the quaternions.