Enumeration via Burnside's Lemma should be feasible if you program efficiently.

$$ |X/G| = \frac 1{|G|} \sum_{g\in G} |X^g|$$

where $X^g$ is the set of fixed points of $g$.

The biggest piece comes from the identity. To enumerate all independent sets, you can use your understanding of the $288$ classes of independent subsets of the $5$-cube. Every independent subset of the $6$-cube induces an independent subset on the top facet and on the bottom facet. So, for each ordered pair $(S_0,S_1)$ of independent subsets of the $5$-cube, determine for each element of the orbit of $S_1$ by the symmetries of the $5$-cube whether this is disjoint from $S_0$. Then multiply by the size of the orbit of $S_0$. This takes about $288^2 2^5 5! 2^5 \approx 2^{33}$ steps, which is not too much.

Each other element of the symmetry group of the $6$-cube acts nontrivially on the cube. The quotient is a smaller graph than the $6$-cube, and most of these quotients fall to a brute force enumeration. Note that if $g$ reverses an edge then you may delete the vertex from the quotient because the edge turns into a loop, and that vertex can't be included in an independent set fixed by $g$.

The largest quotients by nontrivial elements come from the element which transposes the first two coordinates and its conjugates. The quotient has size $48$, and has the structure of a $4$-cube times a chain with $3$ vertices. You can make a $21\times21$ transfer matrix $M$ indexed by the classes of independent sets whose value counts images of the row under the symmetries of the $4$-cube which are disjoint from the column. The number of independent subsets of the quotient equals the $(\emptyset,\emptyset)$ entry of $M^4$.

Similarly, the third largest quotients, by a $3$-cycle, have a quotient of size $32$, which is a little bit too large to brute-force comfortably as a step in a larger computation. However, the quotient is again a product with a chain, and again a transfer matrix simplifies the calculation.

The second largest quotients, by two disjoint transpositions, have the structure of the product of a square with two chains on $3$ vertices. A brute-force check could be done since there are only $36$ vertices, but you can construct a transfer matrix $M$ indexed by independent subsets of a square times a $3$-chain so that the sum of the entries of $M^2$ counts the independent sets. There are fewer than $2^11$ independent subsets of a square times a $3$-chain, so this computation is feasible.

The other quotients are small enough that you can handle them by brute force enumerations, although back-tracking might be faster.

I would test this method on the $5$-cube first, but another check is whether the sum is divisible by $|G| = 2^6 6!$.