Solution:
We count the possible number of colorings. If four colors are used, there are two different colorings that are mirror images of each other, for a total of $2\binom{n}{4}$ colorings. If three colors are used, we choose one color to use twice (which determines the coloring), for a total of $3\binom{n}{3}$ colorings. If two colors are used, we can either choose one of those colors and color three faces with it, or we can color two faces each color, for a total of $3\binom{n}{2}$ colorings. Finally, we can also use only one color, for $\binom{n}{1}$ colorings. This gives a total of
$$2\binom{n}{4} + 3\binom{n}{3} + 3\binom{n}{2} + \binom{n}{1} = \frac{1}{12} n^{2}(n^{2} + 11)$$
colorings. Setting this equal to $n^{3}$, we get the equation $n^{2}(n^{2} + 11) = 12 n^{3}$, or equivalently $n^{2}(n - 1)(n - 11) = 0$, giving the answers $1$ and $11$.