The answer is $1$ in all dimensions, if you mean what I think you mean by "faces."
The underlying graph you're trying to color is the graph whose vertices are the points $(\pm 1, 0, 0, ...), (0, \pm 1, 0, ...), ...$ where a vertex is adjacent to every other vertex except the opposite one (and note that by "vertex" here I mean "face of an $n$-cube"); in other words it is the Turan graph $T_2(2n)$. In particular the subgraph formed by $(1, 0, 0, ...), (0, 1, 0, ...), ...$ is a $K_n$, so we must use each of the $n$ colors exactly once to color each of these vertices. But now it's not hard to see that, since a vertex and its opposite share the same neighbors, they must have the same color. So the only valid colorings are those which use each of the $n$ colors exactly twice each, once for each pair of opposite vertices.
Now every permutation of the set of pairs of opposite vertices is realizable by a rotation, and it follows that the coloring is unique up to rotation.