Here's another approach, one that doesn't involve recursion. It still uses sums and products whose lengths depend on the parameters, though. First I'll give the expression, then explain.
We have
\begin{align}
P &\bigl(
| L_{1} \cap L_{2} \cap \cdots \cap L_{m} | = k
\bigr) \\
&= \frac{\binom{n}{k}}{\prod_{i = 1}^{n} \binom{n}{a_{i}}}
\sum_{j = 0}^{\min(a_{1}, \ldots, a_{m}) - k}
(-1)^{j} \binom{n - k}{j} \prod_{l = 1}^{n} \binom{n - j - k}{a_{l} -j - k}.
\end{align}
EDIT: At the end of writing all of this, I realized that we can consolidate the expression above a little by combining the binomial coefficients into hypergeometric probabilities and trinomial coefficients. For what it's worth, the revised expression is
\begin{equation}
\sum_{j = 0}^{\min(a_{1}, \ldots, a_{m}) - k}
(-1)^{j} \binom{n}{j, k, n - j - k}
\prod_{l = 1}^{n} P( \text{Hyp}(n, j + k, a_{l}) = j + k).
\end{equation}
Here $\text{Hyp}(n, j + k, a_{l})$ is a hypergeometric random variable where $a_{l}$ draws are taken from a population of size $n$ having $j + k$ success states.
Derivation
Let's get some notation in order to make the combinatorial arguments a little easier to track (hopefully). Throughout, we consider $S$ and $a_{1}, \ldots, a_{m}$ fixed. We'll use $\mathcal{C}(I)$ to denote the collection of ordered $m$-tuples $(L_{1}, \ldots, L_{m})$, where each $L_{i} \subseteq S$, satisfying
- $|L_{i}| = a_{i}$; and
- $L_{1} \cap \cdots \cap L_{m} = I$.
We'll also use $\mathcal{C}'(I)$ for a collection identical except that we require $L_{1} \cap \cdots \cap L_{m} \supseteq I$ instead of equality.
A key observation is that $\mathcal{C}'(I)$ is relatively easy to count. This is because the condition $L_{1} \cap \cdots \cap L_{m} \supseteq I$ is equivalent to $L_{i} \supseteq I$ for all $i$, so in a sense this removes interactions between different $i$ values. For each $i$, the number of $L_{i}$ satisfying the requirement is $\binom{|S| - |I|}{a_{i} - |I|}$, since we can construct such an $L_{i}$ by choosing a subset of $S \setminus I$ of size $a_{i} - |I|$ and then unioning with $I$. It follows that
\begin{equation}
| \mathcal{C}'(I) |
= \prod_{i = 1}^{n} \binom{|S| - |I|}{a_{i} - |I|}.
\end{equation}
Now our original probability can be expressed via the $\mathcal{C}$ as follows:
\begin{equation}
P \bigl(
| L_{1} \cap L_{2} \cap \cdots \cap L_{m} | = k
\bigr)
= \frac{ \sum_{I : |I| = k} | \mathcal{C}(I) | }
{ \sum_{\text{all $I \subseteq S$}} | \mathcal{C}(I) | }.
\end{equation}
We can make two simplifications here right away. First, the denominator is the same as
\begin{equation}
| \mathcal{C}'(\emptyset) |
= \prod_{i = 1}^{n} \binom{|S|}{a_{i}}
= \prod_{i = 1}^{n} \binom{n}{a_{i}}.
\end{equation}
Second, a permutation argument shows that $| \mathcal{C}(I) |$ only depends on $I$ through the cardinality $|I|$. Since there are $\binom{n}{k}$ subsets of $S$ having cardinality $k$, it follows that
\begin{equation}
\sum_{I : |I| = k} | \mathcal{C}(I) |
= \binom{n}{k} | \mathcal{C}(I_{0}) |,
\end{equation}
where $I_{0}$ is an arbitrary, fixed subset of $S$ having cardinality $k$.
Taking a step back, we've now reduced the problem to showing that
\begin{equation}
| \mathcal{C}(I_{0}) |
= \sum_{j = 0}^{\min(a_{1}, \ldots, a_{m}) - k}
(-1)^{j} \binom{n - k}{j} \prod_{l = 1}^{n} \binom{n - j - k}{a_{l} - j - k}.
\end{equation}
Let $J_{1}, \ldots, J_{n - k}$ be the distinct subsets of $S$ formed by adding exactly one element to $I_{0}$. Then
\begin{equation}
\mathcal{C}(I_{0})
= \mathcal{C}'(I_{0})
\setminus \biggl(
\bigcup_{i = 1}^{n - k} \mathcal{C}'(J_{i})
\biggr).
\end{equation}
(This is just saying that if $L_{1} \cap \cdots \cap L_{m} = I_{0}$, then $L_{1} \cap \cdots \cap L_{m}$ contains $I_{0}$ but also does not contain any additional element.) We've now transformed the $\mathcal{C}$-counting problem to a $\mathcal{C}'$-counting problem, which we know more how to handle. More specifically, we have
\begin{equation}
| \mathcal{C}(I_{0}) |
= | \mathcal{C}'(I_{0}) |
- \biggl|
\bigcup_{i = 1}^{n - k} \mathcal{C}'(J_{i})
\biggr|
= \prod_{l = 1}^{n} \binom{n - k}{a_{l} - k}
- \biggl|
\bigcup_{i = 1}^{n - k} \mathcal{C}'(J_{i})
\biggr|.
\end{equation}
We can apply inclusion-exclusion to handle the size of the union expression above. The crucial relationship here is that, for any nonempty $\mathcal{I} \subseteq \{ 1, \ldots, n - k \}$,
\begin{equation}
\bigcap_{i \in \mathcal{I}} \mathcal{C}'(J_{i})
= \mathcal{C}' \biggl(
\bigcup_{i \in \mathcal{I}} J_{i}
\biggr).
\end{equation}
This is because if $L_{1} \cap \cdots \cap L_{m}$ contains a number of the $J_{i}$, then it also contains their union. We also note that the set $\bigcup_{i \in \mathcal{I}} J_{i}$ has size $|I_{0}| + |\mathcal{I}| = k + |\mathcal{I}|$. Therefore
\begin{align}
\biggl|
\bigcup_{i = 1}^{n - k} \mathcal{C}'(J_{i})
\biggr|
&= \sum_{\emptyset \neq \mathcal{I} \subseteq \{ 1, \ldots, n - k \}}
(-1)^{| \mathcal{I} | - 1}
\biggl|
\bigcap_{i \in \mathcal{I}} \mathcal{C}'(J_{i})
\biggr| \\
&= \sum_{j = 1}^{n - k}
\sum_{\mathcal{I} : |\mathcal{I}| = j}
(-1)^{j - 1}
\prod_{l = 1}^{n} \binom{n - j - k}{a_{l} - j - k} \\
&= \sum_{j = 1}^{n - k}
(-1)^{j - 1}
\binom{n - k}{j}
\prod_{l = 1}^{n} \binom{n - j - k}{a_{l} - j - k}.
\end{align}
(We can restrict the $j$ values here since the product of the binomial coefficients is zero unless $j \leq a_{l} - k$ for all $l$, i.e. $j \leq \min(a_{1}, \ldots, a_{m}) - k$.)
Finally, by substituting the expression at the end into the equation for $| \mathcal{C}(I_{0}) |$ above and consolidating the sum, we obtain
\begin{equation}
| \mathcal{C}(I_{0}) |
= \sum_{j = 0}^{\min(a_{1}, \ldots, a_{m}) - k}
(-1)^{j} \binom{n - k}{j} \prod_{l = 1}^{n} \binom{n - j - k}{a_{l} - j - k}
\end{equation}
as claimed.
