Number of $2$-colorings of edges of the $n$-dimensional cube?

The calculation for $C(4)$ is actually less complicated than one might think if we build on the results for three dimensions.

The symmetry group of the $n$-dimensional hypercube is the semidirect product of the group $P$ of permutations of the axes (which has $n!$ elements) and the group $R$ generated by the reflections in the axes (which has $2^n$ elements). Here $R$ is the normal subgroup (whereas $P$ is not a normal subgroup of the symmetry group). Thus, we can write elements of the symmetry group as $\pi\sigma$, where $\sigma$ is a product of reflections in the axes (which I’ll denote by a string of $n$ signs indicating the orientations of the axes) and $\pi$ is a permutation of the axes.

An edge is mapped to itself (possibly inverted) by a symmetry element $g=\pi\sigma$ exactly if its direction is a fixed point of the permutation $\pi$ and its position is a fixed point of the restriction of $g$ to the remaining $n-1$ directions. (Note that the first condition ensures that the second condition is well-defined: if a direction is fixed by $\pi$, it forms a one-dimensional invariant subspace of $g$, and the remaining $n-1$ directions also form an invariant subspace, so $g$ can be restricted to them.)

This allows us to construct the edge cycle index monomials of a symmetry element that leaves at least one axis fixed (possibly inverted) from the edge and vertex cycle index monomials of its restriction to the remaining $n-1$ directions; so the only work left to do for $n$ is for the symmetry elements whose permutations have no fixed points, and those turn out to be relatively easy to handle.

So let’s work out the cycle index monomials for the full octahedral group (in $3$ dimensions) for edges and vertices as a basis for finding the edge cycle index of the hyperoctahedral group for $n=4$. I’ll just write them out in a table without going into the details of how to obtain them since you wrote that this part seems intuitive to you. The table only includes one representative $\pi$ for each conjugacy class of $S_3$ and one representative $\sigma$ for each class of equivalent signatures (depending on $\pi$); the last column counts the symmetry elements corresponding to these classes.

\begin{array}{c|c} \pi&\sigma&\text{edges}&\text{vertices}&\text{count}\\\hline (1)(2)(3)&+++&a_1^{12}&a_1^8&1\\ &++-&a_1^4a_2^4&a_2^4&3\\ &+--&a_2^6&a_2^4&3\\ &---&a_2^6&a_2^4&1\\ (12)(3)&+++&a_1^2a_2^5&a_1^4a_2^2&3\\ &++-&a_1^2a_2^5&a_2^4&3\\ &+-+&a_4^3&a_4^2&6\\ &+--&a_4^3&a_4^2&6\\ &--+&a_1^2a_2^5&a_1^4a_2^2&3\\ &---&a_1^2a_2^5&a_2^4&3\\ (123)&+++&a_3^4&a_1^2a_3^2&2\\ &++-&a_6^2&a_2^1a_6^1&6\\ &+--&a_3^4&a_1^2a_3^2&6\\ &---&a_6^2&a_2^1a_6^1&2 \end{array}

Now, if the permutation of a symmetry element $g$ fixes a direction, the edges in that direction form exactly the cycles that the corresponding vertices form under the restriction of $g$ to the remaining $n-1$ directions; so one factor in the edge cycle index monomial for $n$ is the vertex cycle index monomial for $n-1$. If the fixed direction is not reflected, the edges in the remaining $n-1$ directions form the same cycles as under the restriction, but there are twice as many of them, so another factor in the edge cycle index monomial for $n$ is the square of the edge cycle index monomial for $n-1$. If the fixed direction is reflected, then cycles of even length under the restriction retain their length, but there are twice as many, so these factors get squared; whereas cycles of odd length are joined in pairs to form cycles of twice the length, so in these factors, $a_k$ is replaced by $a_{2k}$.

This allows us to obtain the following edge cycle index monomials for $n=4$, building on the table above:

\begin{array}{c|c} \pi&\sigma_1\sigma_2\sigma_3&\sigma_4=+&\sigma_4=-&\text{count}\\\hline (1)(2)(3)(4)&+++&a_1^{32}&a_1^8a_2^{12}&1\\ &++-&a_1^8a_2^{12}&a_2^{16}&3\\ &+--&a_2^{16}&a_2^{16}&3\\ &---&a_2^{16}&a_2^{16}&1\\ (12)(3)(4)&+++&a_1^8a_2^{12}&a_1^4a_2^{14}&6\\ &++-&a_1^4a_2^{14}&a_2^{16}&6\\ &+-+&a_4^8&a_4^8&12\\ &+--&a_4^8&a_4^8&12\\ &--+&a_1^8a_2^{12}&a_1^4a_2^{14}&6\\ &---&a_1^4a_2^{14}&a_2^{16}&6\\ (123)(4)&+++&a_1^2a_3^{10}&a_1^2a_3^2a_6^4&8\\ &++-&a_2^1a_6^5&a_2^1a_6^5&24\\ &+--&a_1^2a_3^{10}&a_1^2a_3^2a_6^4&24\\ &---&a_2^1a_6^5&a_2^1a_6^5&8 \end{array}

We’re fortunate that these are the monomials that would have been more difficult to work out in $4$ dimensions directly, whereas the remaining symmetry elements whose permutations don’t have fixed points are relatively easy to handle. If we apply such a symmetry element repeatedly, then, since all cycles in the permutation are of length at least $2$, the edges cannot return to their original position before we get back to the identity permutation. At this point, an axis is reflected exactly if the permutation cycle in which it lies contains an odd number of reflections. Again, since there are no cycles of length $1$, if some axis is reflected, then at least two axes are reflected, so again no edge can be in its original position. Thus, all edges form cycles of the same length, which is either the order of the permutation if all cycles have even numbers of reflections, or twice that if at least one cycle has an odd number of reflections.

Thus we obtain the following monomials for the derangements of the axes:

\begin{array}{c|c} \pi&\text{$\exists$ odd}&\text{monomial}&\text{count}\\\hline (12)(34)&\text{no}&a_2^{16}&12\\ &\text{yes}&a_4^8&36\\ (1234)&\text{no}&a_4^8&24\\ &\text{yes}&a_8^4&24 \end{array}

Now all that remains is to add up the monomials (weighted by the counts) to obtain the edge cycle index of the hyperoctahedral group for $n=4$, substitute $2$ for all variables and divide by the number $2^4\cdot4!=384$ of symmetry elements. The result is

$$ \frac{1\cdot2^{32}+16\cdot2^{20}+24\cdot2^{18}+35\cdot2^{16}+32\cdot2^{12}+164\cdot2^8+64\cdot2^6+48\cdot2^4}{384}=\boxed{11251322}\;. $$

Here’s Java code that computes the number of equivalence classes of $2$-colorings of the edges of the $n$-dimensional hypercube under rotations and reflections (by counting the edge cycles of each symmetry element); the results up to $n=8$ are (in agreement with the above calculation):

1 : 2
2 : 6
3 : 144
4 : 11251322
5 : 314824456456819827136
6 : 136221825854745676520058554256163406987047485113810944
7 : 1126672129674489847701704811334332425523379727144553194843038610078657640531358451246775872508990558612282358941688264175298543616
8 : 17416266885059329153193448416467331016109182971213752603530165042718086926221183959254526030274624207791564091034795976387518055177618065547557580558563049317223935447616138542363134382681407634060633896718472077319869457188945414864160284255850798170929140736717375938929354322271320922284872609824768