Here's an explanation in terms of the Hodge dual and the exterior (wedge) product.
Let ${e_1, e_2, e_3}$ be the standard orthonormal basis for $\mathbb{R}^3$. Consider the two vectors $a = a_1 e_1 + a_2 e_2 + a_3 e_3$ and $b = b_1 e_1 + b_2 e_2 + b_3 e_3$. From the matrix computation we obtain the familiar formula
$a\times b = (a_2 b_3 - a_3 b_2) e_1 + (a_3 b_1 - a_1 b_3) e_2 + (a_1 b_2 - a_2 b_1) e_3$.
But (see note at the bottom)
$a \wedge b = (a_1 b_2 - a_2 b_1) e_1 \wedge e_2 + (a_2 b_3 - a_3 b_2) e_2 \wedge e_3 + (a_3 b_1 - a_1 b_3) e_3 \wedge e_1$,
where the wedge $\wedge$ represents the exterior product. One can now compute the dual of this latter expression using that the left contraction of $(e_1 \wedge e_2)$ onto $(e_3 \wedge e_2 \wedge e_1)$ is $e_3$ (and similar relations). The result is that
$a \times b = (a \wedge b)^*$,
that is, the cross product of $a$ and $b$ is the dual of their exterior product.
Geometrically, this is an incredible picture. The exterior product is the plane element spanned by both $a$ and $b$, and the dual is the vector orthogonal to that plane.
This is my favorite interpretation of the cross product, but it's only helpful, of course, if you're familiar with exterior algebra and the Hodge dual.
Note: The wedge product can be found by formally computing
$(a_1 e_1 + a_2 e_2 + a_3 e_3) \wedge (b_1 e_1 + b_2 e_2 + b_3 e_3)$
using the distributivity and anticommutation relations of the exterior product.
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.
Best Answer
One point of view is that the cross product is the composition of the Hodge dual and the exterior product $V \times V \to \Lambda^2 V$ in three dimensions. The Hodge dual requires additional structure to define: you need not only an inner product, but an orientation. This reflects the fact that there is a choice of handedness in the definition of the cross product.
Another point of view is that the cross product is not an operation on spatial vectors in $\mathbb{R}^3$, but the Lie bracket on the Lie algebra of $\text{SO}(3)$. This is the definition, for example, which is relevant to the physics of angular momentum. Of course, if you insist on taking the cross product of spatial vectors you need a way to identify spatial vectors with elements of the Lie algebra of $\text{SO}(3)$. Writing $\text{SO}(3)$ as $\text{Aut}(V)$ where $V$ is a real oriented 3-dimensional inner product space, the Lie algebra of $\text{SO}(3)$ is naturally isomorphic to $\Lambda^2 V$, so this identification is again the Hodge dual.