The required number is $m = |S| - n$. We begin by showing that any set $\mathcal{A}$ of subsets of $S$ satisfying the two conditions in the statement has a subcover of cardinality at most $m$.

This is clear if $S$ is a member of $\mathcal{A}$.

Assume henceforth that $\mathcal{A}$ does not contain $S$. If some member $A$ of $\mathcal{A}$ has more than $n$ elements, for each element of $S \setminus A$ choose a containing member of $\mathcal{A}$. The latter along with $A$ form a subcover of $\mathcal{A}$ of cardinality $|S \setminus A| + 1 \le |S| - (n+1) + 1 = m$.

Assume henceforth that each member of $\mathcal{A}$ has exactly $n$ elements. Fix a member $A$ of $\mathcal{A}$.

If some member $B$ of $\mathcal{A}$ contains more than one element of $S \setminus A$, for each element of $S \setminus (A \cup B)$ choose a containing member of $\mathcal{A}$. The latter along with $A$ and $B$ form a subcover of $\mathcal{A}$ of cardinality $|S \setminus (A \cup B)| + 2 = |S \setminus A| - |(S \setminus A) \cap B| + 2 \le |S \setminus A| = m$.

Finally, if no member of $\mathcal{A}$ contains more than one element of $S \setminus A$, write $S \setminus A = \{x_1, x_2, \dots, x_m\}$, choose a member $A_1$ of $\mathcal{A}$ containing $x_1$ and notice that $A \setminus A_1$ is a singleton set, say $A \setminus A_1 = \{x\}$. Since $x_2$ is contained in at least $n$ members of $\mathcal{A}$, each of which contains (exactly) $n-1$ elements of $A$, we may choose a member $A_2$ of $\mathcal{A}$ containing both $x$ and $x_2$ (recall that $n \ge 2$). If $n \ge 3$, continue choosing members $A_i$ of $\mathcal{A}$ containing $x_i$, $i = 3, \dots, n$, to form an $m$-element subcover of $\mathcal{A}$ consisting of $A_1, A_2, \dots, A_m$.

To complete the proof, we produce a set of subsets of $S$ satisfying the two conditions in the statement, no subcover of which has less than $m$ members. To this end, write $S = \{1, 2, \dots, m+n\}$, $m \ge 2$, and let $S_1, S_2, \dots, S_n$ be the $(n-1)$-element subsets of the upper part $\{m+1, m+2, \dots, m+n\}$ of $S$. The sets $S_{i,j} = S_j \cup \{i\}$, $i = 1, 2, \dots, m$, $j = 1, 2, \dots, n$, satisfy both conditions in the statement and at least $m$ of them are needed to cover $S$. (The condition $m \ge 2$ is required for an element in the upper part to lie in at least $n$ of these sets.)