First let's state the special case of Burnside's lemma that is relevant here.
Lemma: Let $G$ be a finite group acting on a finite set $X$. The number of ways to color the elements of $X$ with $z$ different colors, up to the action of $G$, is
$$\frac{1}{|G|} \sum_{g \in G} z^{c(g)}$$
where $c(g)$ is the number of cycles in the cycle decomposition of $g$ acting on $X$. (Proof.)
Here $X$ is the set of faces of a hypercube. In $n$ dimensions there are $2n$ such faces. $G$ is the subgroup of index $2$ in the hyperoctahedral group with determinant $1$, also known as the Coxeter group $D_n$. So our job now is to count, for each $k$, the number of elements of $D_n$ with $k$ cycles in the action on $X$.
Now, note that to analyze the action of $D_n$ on the faces it suffices to analyze the action of $D_n$ on the midpoints of the faces. But these are precisely the $2n$ points $(0, 0, ... \pm 1, ..., 0, 0)$, so writing the elements of $D_n$ as signed permutation matrices is very well-suited to analyzing their action on these points; in particular, it suffices to figure out the answer for a single signed cycle. But this turns out to be very simple: there are either one or two cycles depending on whether the product of the signs is $-1$ or $+1$.
(It might be helpful here to play with a specific example. Consider $\left[ \begin{array}{cc} 0 & 1 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & 0 \end{array} \right]$ acting on the six points $(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1)$ to get a feel for what's going on in the general case.)
From here I think it's easiest to work with generating functions because the combinatorics get a little messy. Begin with the identity
$$\sum_{m \ge 0} Z(S_n) t^n = \exp \left( z_1 t + \frac{z_2 t^2}{2} + \frac{z_3 t^3}{3} + ... \right)$$
where $Z(S_n)$ is the cycle index polynomial for the action of $S_n$ on $\{ 1, 2, ... n \}$. Each $z_i$ is the term that controls cycles of length $i$. We want to modify this generating function so that it tells us how the cycles in $D_n$ work. There are $2^i$ signed cycles of length $i$ which come in two flavors: half of them have positive sign product (two unsigned cycles) and half of them have negative sign product (one unsigned cycle), so to keep track of the total number of unsigned cycles we should replace $z_i$ with $2^{i-1} z^2 + 2^{i-1} z$. We also have to keep in mind that the determinant of a signed cycle is its sign product multiplied by $(-1)^{i+1}$, and we only want permutations with determinant $1$. So the generating function we want is
$$\sum_{n \ge 0} f_n(z) \frac{t^n}{n!} = \frac{1}{2} \left( \exp \left( \sum_{i \ge 1} \frac{2^{i-1} z^2 + 2^{i-1} z) t^i}{i} \right) + \exp \left( \sum_{i \ge 1} (-1)^{i+1} \frac{2^{i-1} z^2 - 2^{i-1} z) t^i}{i} \right) \right)$$
where $f_n(z) = \sum_{g \in D_n} z^{c(g)}$. After some simplification the above becomes
$$\sum_{n \ge 0} \frac{1}{|D_n|} f_n(z) t^n = \frac{1}{(1 - t)^{(z^2+z)/2}} + (1+t)^{(z^2-z)/2}.$$
Substituting $z = 2$ gives, at last, the answer
$$\sum_{n \ge 0} \frac{1}{|D_n|} f_n(2) t^n = \frac{1}{(1 - t)^3} + 1 + t.$$
In other words, for $n \ge 2$ we simply have $\frac{1}{|D_n|} f_n(2) = {n+2 \choose 2}$. This is such a simple answer that there should be a direct proof of it. I'll keep working on it. (There is a rather straightforward proof of the corresponding result with "hypercube" replaced by "simplex," so my guess is something along those lines is possible here.)
Here are a couple hints. I will consider the vertices to be length-$n$ bitstrings.
An edge connects two vertices if the vertices differ by a single bit flip. How many edges are incident to a given vertex? If you multiply this by the number of vertices, what would you do next to end up with the total number of edges?
A facet (i.e. $(n-1)$-dimensional cell) may be chosen by first picking a coordinate, fixing its bit, and then letting all the other bits be arbitrary. How many ways are there to do this?
Best Answer
Consider the $n$-dimension hypercube with vertices at $(\pm x,\pm x,\ldots,\pm x)$. Its volume $V_n(x)$ is $(2x)^n$. Imagine we expand the hypercube a little bit from $x$ to $x + \delta$, each $m$-dimension face of the hypercube will contribute an extra hypervolume $V_m(x) \delta^{n-m}$ to the new hypercube. This means:
$$V_{n}(x+\delta) = \sum_{m=0}^n N_{n,m} V_{m}(x)\delta^{n-m}$$ where $N_{n,m}$ is the number of $m$-dimension "faces" of a $n$-dimension hypercube.
One the other hand, binomial theorem tell us:
$$V_n(x+\delta) = 2^n(x+\delta)^n = 2^n\sum_{m=0}^n \binom{n}{m} x^{m} \delta^{n-m} =\sum_{m=0}^n 2^{n-m}\binom{n}{m}V_{m}(x)\delta^{n-m} $$
By comparing the coefficients of powers in $\delta$, we immediately get:
$$N_{n,m} = 2^{n-m} \binom{n}{m}$$
In particular,
$$N_{n,n-2} = 2^2 \binom{n}{n-2} = 2^2 \binom{n}{2} = 2n(n-1)$$
Update
Let $N_{n,m}^c$ be the number of $m$-dimension face in contact to a corner in a $n$-dimension hypercube. If you compare the volume of the hypercube $[0,x]^n$ with that of $[0,x+\delta]^n$ like above, you will obtain:
$$N_{n,m}^c = \binom{n}{m}$$
An alternate combinatorial argument go like this. Since a $n$-dimension hypercube has $N_{n,0} = 2^n$ vertices. There are $2^n N_{n,m}^c$ ways of picking a corner and then a $m$-dimension face in contact to that corner. We can arrive the same count by picking the $m$-dimension face first. As a result, we again get:
$$2^n N_{n,m}^c = 2^m N_{n,m} \quad\implies\quad N_{n,m}^{c} = 2^{m-n} N_{n,m} = \binom{n}{m}$$
BTW, you $n$ choose $2$ argument also works. In any event,
$$N_{n,n-2}^c = \binom{n}{n-2} = \binom{n}{2} = \frac{n(n-1)}{2}$$
For tesseract, this is $6 \ne 2 n - 3$. Let's say your tesseract is $\{\;(x,y,z,t) : 0\le x,y,z,t \le 1\;\}$. The six $2$-dimension faces in contact to the origin are:
$$\begin{cases} x = y = 0\text{ face} &\to&\{0\} &\times& \{0\} &\times& [0,1] &\times& [0,1]\\ x = z = 0\text{ face} &\to&\{0\} &\times& [0,1] &\times& \{0\} &\times& [0,1]\\ x = t = 0\text{ face} &\to&\{0\} &\times& [0,1] &\times& [0,1] &\times& \{0\}\\ y = z = 0\text{ face} &\to&[0,1] &\times& \{0\} &\times& \{0\} &\times& [0,1]\\ y = t = 0\text{ face} &\to&[0,1] &\times& \{0\} &\times& [0,1] &\times& \{0\}\\ z = t = 0\text{ face} &\to&[0,1] &\times& [0,1] &\times& \{0\} &\times& \{0\} \end{cases}$$