How many strings are close to a given set of strings? This question has been prompted by Efficient data structures for building a fast spell checker.
Given two strings $u,v$, we say they are $k$-close if their Damerau–Levenshtein distance¹ is small, i.e. $\operatorname{LD}(u,v) \geq k$ for a fixed $k \in \mathbb{N}$. Informally, $\operatorname{LD}(u,v)$ is the minimum number of deletion, insertion, substitution and (neighbour) swap operations needed to transform $u$ into $v$. It can be computed in $\Theta(|u|\cdot|v|)$ by dynamic programming. Note that $\operatorname{LD}$ is a metric, that is in particular symmetric.
The question of interest is:

Given a set $S$ of $n$ strings over $\Sigma$ with lengths at most $m$, what is the cardinality of
$\qquad \displaystyle S_k := \{ w \in \Sigma^* \mid \exists v \in S.\ \operatorname{LD}(v,w) \leq k \}$?

As even two strings of the same length have different numbers of $k$-close strings² a general formula/approach may be hard (impossible?) to find. Therefore, we might have to compute the number explicitly for every given $S$, leading us to the main question:

What is the (time) complexity of finding the cardinality of the set $\{w\}_k$ for (arbitrary) $w \in \Sigma^*$?

Note that the desired quantity is exponential in $|w|$, so explicit enumeration is not desirable. An efficient algorithm would be great.
If it helps, it can be assumed that we have indeed a (large) set $S$ of strings, that is we solve the first highlighted question.


Possible variants include using the Levenshtein distance instead.
Consider $aa$ and $ab$. The sets of $1$-close strings over $\{a,b\}$ are $\{ a, aa,ab,ba,aaa,baa,aba,aab \}$ (8 words) and $\{a,b,aa,bb,ab,ba,aab,bab,abb,aba\}$ (10 words), respectively .
See Levenshtein's paper. It contains bounds on the number strings obtained from insertion and deletion of a string. If $n$ is the length of the string and the string is binary, then the maximum number of nearest neighbors in the Levenshtein distance is $\Theta(n^2)$. It is comparatively harder to say anything about $k$-nearest neighbours, but one can get bounds. These should give you an estimate on the complexity.