This is not a complete answer and may not tell you anything you don't already know, but it's too long for a comment.

You can get an estimate on the number of words at an edit distance of $k$ from a given word of length $n$ by using the arguments in the proof of Lemma 2.6 in this paper. (The lemma is on page 10 and its proof is on page 28.)

The idea is that if a word $w$ is at an edit distance of $k$ from a word $v$, then one can get from $v$ to $w$ via the following steps:

- insert $k$ copies of the symbol 'e' (for 'edit') into the word $v$;
- one by one, go through the symbols 'e' and either change the symbol immediately before it, delete the symbol immediately before it, or insert a symbol immediately before it, and then delete the 'e'.

Step 1 gives a word of length $n+k$ with $k$ occurrences of the symbol 'e', so there are ${n+k\choose k}$ possibilities after this step. Then step 2 gives at most $3^k$ possible words for each of those possibilities, so for a fixed word $v$ of length $n$ one obtains the bound
$$
\#\{w \mid \text{edit distance from $v$ to $w$ is $k$}\} \leq 3^k {n+k \choose k}.
$$
For small $k$ this is a reasonable bound; in particular when $k\ll n$ this is not much larger than ${n\choose k}$, the number of words a Hamming distance of $k$ away. The problem is that for larger values of $k$ the procedure described above succumbs to a lot of overcounting, so it's not clear what the actual bound should be.