There is an algorithmic approach to this which I include for future
reference and which consists in using Burnside and Stirling numbers of
the second kind. For Burnside we need the cycle index of the face
permutation group of the cube. We enumerate the constituent
permutations in turn. First there is the identity for a contribution
of $$a_1^6.$$
Rotating about one of the four diagonals by $120$ degrees and $240$
degrees we get
$$4\times 2a_3^2.$$
Rotating about an axis passing through opposite faces by $90$ degrees
and by $270$ degrees we get
$$3\times 2 a_1^2 a_4$$
and by $180$ degrees
$$3\times a_1^2 a_2^2.$$
Finally rotating about an exis passing through opposite edges yields
$$6\times a_2^3.$$
We thus get the cycle index
$$Z(G) = \frac{1}{24}
(a_1^6 + 8 a_3^2 + 6 a_1^2 a_4 + 3 a_1^2 a_2^2 + 6 a_2^3).$$
As a sanity check we use this to compute the number of colorings with
at most $N$ colors and obtain
$$\frac{1}{24}(N^6 + 8 N^2 + 12 N^3 + 3 N^4).$$
This gives the sequence
$$1, 10, 57, 240, 800, 2226, 5390, 11712, 23355, 43450, \ldots$$
which is OEIS A047780 which looks to be
correct. Now if we are coloring with $M$ colors where all $M$ colors
have to be present we must partition the cycles of the entries of the
cycle index into a set partition of $M$ non-empty sets. We thus obtain
$$\frac{M!}{24}
\left({6\brace M} + 8 {2\brace M} + 12 {3\brace M}
+ 3{4\brace M}\right).$$
This yields the finite sequence (finite because the cube can be
painted with at most six different colors):
$$1, 8, 30, 68, 75, 30, 0, \ldots$$
In particular the value for four colors is $68.$ We also get $6!/24 =
30$ for six colors because all orbits have the same size.
