Let $a_n$ be defined by the recursion
$$a_{n+2} + 4 a_{n+1} + a_n =0,\ a_1=-4,\ a_2=15.$$
Let $b_n$ be defined by the recursion
$$b_{n+2} + 6 b_{n+1} + b_n =0,\ b_1 =-6,\ b_2=35.$$

The number of unfoldings of the labeled $1 \times 1 \times n$ cube is $-4 a_n^2 b_n$. (I might have a sign error here.)

We count spanning trees using the matrix tree theorem. Let $G$ be the dual graph to the $1 \times 1 \times n$ cube (so $G$ has $4n+2$ vertices). Let $V$ be the vector space of functions on the vertices of $G$. Let $\Lambda: V \to V$ be the Laplacian operator. Then $\Lambda$ has a one dimensional kernel -- the constant function -- and we are supposed to compute the coefficient of $t$ in $\det(\Lambda - t \mathrm{Id})$.

The cyclic group of order $4$ acts on the $1 \times 1 \times n$ cube by rotation around the long axis. This action commutes with $\Lambda$. So we can compute $\Lambda$ separately on each of the four eigenspaces of this rotation. The corresponding matrices are
$$\begin{pmatrix}
-4 & 1 & 0 & 0 & 0 & \cdots & 0 \\
1 &-4 & 1 & 0 & 0 & \cdots & 0 \\
0 & 1 &-4 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 &-4 & 1 & \cdots & 0 \\
0 & 0 & 0 & 1 &-4 & \cdots & 0 \\
& & & & & \ddots & \\
0 & 0 & 0 & 0 & 0 & \cdots & -4 \end{pmatrix} \ \mathrm{on\ the} \pm i \ \mathrm{eigenspace}$$
$$\begin{pmatrix}
-6 & 1 & 0 & 0 & 0 & \cdots & 0 \\
1 &-6 & 1 & 0 & 0 & \cdots & 0 \\
0 & 1 &-6 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 &-6 & 1 & \cdots & 0 \\
0 & 0 & 0 & 1 &-6 & \cdots & 0 \\
& & & & & \ddots & \\
0 & 0 & 0 & 0 & 0 & \cdots & -6 \end{pmatrix} \ \mathrm{on\ the} -1 \ \mathrm{eigenspace}$$
$$\begin{pmatrix}
-4 & 4 & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\
1 & -2 & 1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0\\
0 & 1 &-2 & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\
0 & 0 & 1 &-2 & 1 & 0 & \cdots & 0 & 0 & 0\\
0 & 0 & 0 & 1 &-2 & 1 & \cdots & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 &-2 & \cdots & 0 & 0 & 0\\
& & & & & \ddots & \\
0 & 0 & 0 & 0 & 0 & 0 & \cdots & 1 & -2 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 4 & -4 \end{pmatrix} \ \mathrm{on\ the}\ 1 \ \mathrm{eigenspace}$$

The first two matrices are invertible and obey the recursions $a_n$ and $b_n$ respectfully. (See the recursive formula displayed in this article.)

The last matrix has a one dimension kernel. Taking the cofactor by deleting the last row and column, one gets a matrix whose determinant is $-4$, as can be checked from the same recursion as above.

Putting it all together, the number of spanning trees is $-4 a_n^2 b_n$.

I don't understand the statistical question you want to ask about the unfolded shapes.

**Remark:** Using the cyclic symmetry broght this computation down into the range of reasonable computation. But, even without that, I immediately knew that the answer would obey some linear recursion. Consider building a spanning tree of $G$ one layer at a time as we travel along the long axis of the cube. At the $k$-th partial stage, there are $4$ vertices of $G$ on the exposed boundary; call them $u_k$, $v_w$, $w_v$, $x_k$. The graph so far is a forest, each connected component of which contains at least one of the four exposed vertices. There are $15$ ways to group these $4$ vertices into connected components, one of which is actually impossible for planarity reasons. So there is some $14 \times 14$ matrix which records, for example, if the vertices on one level are grouped as $(\{ u_k, v_k \}, \{ w_k \}, \{ x_k \})$, how many ways are there to extend the forest to one where the vertices on the next level are grouped as $( \{ u_{k+1} \}, \{ v_{k+1}, w_{k+1}, x_{k+1} \})$. Taking powers of this $14 \times 14$ matrix and then pairing with some row and column vectors to handle end effects, you get this count.

This is what I learned to call the "transfer matrix method" (although Wikipedia seems to call something else by that name) and it solves almost all "linear" combinatorics problems.