Since you want the length to remain unchanged and $2$ to be the minimal edit distance, the only options are two substitutions in different places, or an insertion and a deletion. (It doesn't matter in which order we carry out the insertion and the deletion.) It's straightforward that there are $\binom n2=\frac{n(n-1)}2$ different results of two substitutions in different places, so the task is to count the strings produced by an insertion and a deletion that can't be produced by at most two substitutions.
Let's count the cases where the insertion is to the left of the deletion and then multiply by $2$. The combined effect of the insertion and the deletion is to shift all $k$ bits between them to the right while replacing the first one and removing the last one. This result can also be achieved by at most $k$ substitutions, so we need $k\gt2$. Inserting $x$ within a run of $x$s has the same effect as inserting $x$ at the end of the run. Thus we can count all insertions with different effects once by always inserting the bit complementary to the one to the right of the insertion. Similarly, a deletion within a run has the same effect as a deletion at the start of the run, so we should only count deletions that follow a change between $0$ and $1$.
That gives us an initial count of
$$
2\cdot\frac12\sum_{k=3}^n(n+1-k)=\sum_{k=1}^{n-2}k=\frac{(n-1)(n-2)}2\;,
$$
which together with $\frac{n(n-1)}2$ from the substitutions yields $(n-1)^2$. That's already of the order of the counts you computed, but a bit too high, so we're overcounting.
If there are no further changes in the $k$ shifted bits other than the one preceding the deletion, then only the bits next to the insertion and deletion change, and we can achieve that with $2$ substitutions, so we have to subtract
$$
\sum_{k=3}^n\left(\frac12\right)^{k-2}(n+1-k)=\sum_{k=1}^{n-2}\left(\frac12\right)^{n-k-1}k=n-3+2^{-(n-2)}\;.
$$
Also, if the entire range of shifted bits consists of alternating zeros and ones, then swapping the insertion and the deletion yields the same effect, so in this case we were double-counting and need to subtract
$$
\sum_{k=3}^n\left(\frac12\right)^{k-1}(n+1-k)\;,
$$
which is half the previous sum. Thus, the expected number of binary strings of length $n$ at edit distance exactly $2$ from a uniformly randomly selected binary string of length $n$ is
$$
(n-1)^2-\frac32\left(n-3+2^{-(n-2)}\right)=n^2-\frac72n+\frac{11}2-6\cdot2^{-n}\;,
$$
in agreement with your computed results.
Actually this seems quite easy
Let $A^d_{i,j}$ be the event that two strings $s_i$ $s_j$
have distance $d$. Then, if both $s_i,s_j$ are random, we have $$P(A^d_{i,j})= \binom{n}{d} 2^{-n}$$
Then, if we sample $k$ random strings, the expected number of pairs with such distance is
$$ \frac{k(k-1)}{2} \binom{n}{d} 2^{-n}$$
Best Answer
Your formulation in terms of Hamming distance seems to be a good one for making progress.
The quantity you want then seems to be denoted $A_2(n,n/2)$. The Plotkin bound gives: if $n/2$ is even then $A_2(n,n/2) \le 2n$, otherwise $A_2(n,n/2) \le 2\lfloor n/2+1 \rfloor = n+1$.
A lower bound can be found by constructing an explicit set. For the special case $n=2^t$ for positive integer $t$, consider the strings $0^n, 0^{n/2}1^{n/2}, 0^{n/4}1^{n/4}0^{n/4}1^{n/4},\dots, 0^21^20^2\dots 0^21^2, 0101\dots 01$. This satisfies the conditions and has $t+1 = 1 + \log_2 n$ strings.
I suspect that better bounds are known, and would be interested in a more precise analysis. Perhaps someone like Peter Cameron might be able to provide pointers.
It is worth remarking that there is an interesting reformulation of your problem in terms of set systems.
Given a set of strings which are all distance $n/2$ from each other, first transform the coordinate system. Pick one string in the set and flip its 1 bits to make it the all-zero string. Applying the same permutation to each of the other strings retains the Hamming distances. This leads to every string other than $0^n$ having exactly $n/2$ bits set to 1.
Consider the set system of which these strings are the characteristic vectors of sets in the system, excluding the all-zero vector. Each set contains precisely $n/2$ elements. Observe that the pairwise intersections contain precisely $n/4$ elements.
Then your question can be rephrased as follows (equivalent up to the difference of 1 from excluding the $0^n$ string).
Again this sounds natural and I would expect this is known, but I did not find anything about this problem in the Lovász Combinatorial Problems and Exercises book or in the Deza/Frankl survey from 1983; it seems a bit tricky to find the right online search terms. The usual setup for Erdős–Ko–Rado style results is more relaxed, allowing intersections of at least some threshold; here all pairwise intersections are precisely $n/4$.
There is another potentially relevant paper but I have not been able to extract much from it (or at least from the scanned image of the paper).