Following up on the comments about using spanning trees and automorphism groups. It does seem to me that perhaps the best way to do this would be to compute the number of different spanning trees of the hyperoctahedral graph up to symmetries of the hyperoctahedron. As Peter Taylor and Brendan McKay have noted, one can use Burnside's lemma to do this. I think this will be easier, because it seems likely that about half of all symmetries will not fix any spanning trees (see below), and that most spanning trees are not fixed by any symmetry. If this is true, most of the computation would be finding a small number of spanning trees that are fixed by reflections of the hyperoctohedron. On the other hand, it seems like computing the number of trees with an appropriate pairing would require solving several instances of the graph isomorphism problem to make sure you aren't overcounting.
The rest of this post is not about computation, but rather about general conjectures. I think (tho I haven't sat down to prove) that rotations will never fix a spanning tree. If this is so, then, after counting the number of spanning trees of the hyperoctohedral graph (which can be done using the matrix-tree theorem), we could use a very simple version of Burnside's lemma to get bounds on the number of nets. Specifically, let $N_d$ be the number of nets of the hypercube in $\mathbf{R}^d$, and let $F(d) = \frac{(2d)^{d-2} (d-1)^d}{d!}$ (this is the same number that Brendan mentions). Then, assuming (as I believe) that no rotation fixes a spanning tree, we get that
$F(d) \leq N_d \leq 2F(d).$
For some small $d$, the bounds look like this:
| $d$ | Bounds |
|---|---|
| 2 | $\frac{1}{2} \leq 1 \leq 1$ |
| 3 | $8 \leq 11 \leq 16$ |
| 4 | $216 \leq 261 \leq 432$ |
| 5 | $8533\frac{1}{3} \leq 9694 \leq 17066\frac{2}{3}$ |
The table makes it look like the lower bound is probably closer to the truth. Looking at the ratios $N_d/F(d)$ for the numbers you've computed, Brendan says they appear to be going to 1. I'm a little less sure, since it looks like the rate at which they decrease is going down as $d$ increases, but with such a small dataset it's hard to tell. I'd guess that $N_d/F(d)$ approaches a constant near $1.08$. If it turns out I'm wrong, at least I'll have Legendre to keep me company.