I agree with your formula. You choose $j \in \{0,\ldots,m-\ell\}$ (it will be the cardinal of $A \cap Y$, and given such a $j$ you choose independently $j$ elements in $Y$ and $m-j$ elements in $X$.

Here is an upper bound

$$S_{\ell,m,n} \le \sum_{j=0}^{m} {n \choose m-j} {n \choose j} = {2n \choose m}.$$
One obtains the last equality by looking at the coefficient of $X^m$ in the product $(1+X)^n \times (1+X)^n$. My guess is that this bound is sharp when $m/2-\ell >> \sqrt{m}$.

Less trivial bound

The quotient $S_{\ell,m,n}/{2n \choose m}$ is the probability $\mathbf{P}[X \ge \ell]$ where $X$ is a random hypergeometric random variable with parameters $2n$, $n$ and $m$. The law of $X$ is symmetric with regard to $m/2$ and is less spread out than the binomial law with parameters $m$ and $1/2$ (sampling without replacement provides a least dispersion of the number of success).

More precisely, given binomial random variable $Y$ with parameters $m$ and $1/2$, I guess that when $\ell \ge m/2$,
$$2\mathbb{P}[X \ge \ell] = \mathbb{P}[|X-m/2| \ge \ell-m/2] \le \mathbb{P}[|Y-m/2| \ge \ell-m/2] = 2\mathbb{P}[Y \ge \ell].$$
Does anyone have a reference or a proof of this fact? Then, Cramer-Chernoff inequalities give nice bounds.