Here is a solution to Brendan McKay's more general question, which I'll do as a separate answer.

Let $n_1,n_2,\ldots,n_k$ be integers, with $1 < n_1 \le n_2 \le \cdots \le n_k$, let $n = n_1+n_2 + \cdots +n_k$, and suppose that $G = S_{n_1} \times S_{n_2} \times \cdots S_{n_k} \le S_p$ for some $p$. Then $n \le p$.

To make the inductive argument work, I found that I needed to prove this also for the alternating groups, but then we need degree greater than $2$, because $A_2$ is trivial. So, for $1 \le i \le k$, let $G_i = S_{n_i}$ or (if $n_i>2$) $A_{n_i}$, and assume that $G = G_1 \times \cdots \times G_k \le S_p$, Then $n\le p$, where $n = n_1+n_2 + \cdots +n_k$.

Suppose not, and choose $p$ minimal such that there is a counterexample. Clearly $k > 1$.

If $G$ is intransitive (as a subgroup of $S_p$), then $p = p_1 + p_2$ for some $p_1,p_2 >0$, and $G$ fixes sets of sizes $p_1$ and $p_2$. By induction, those $G_{i}$ that act faithfully on first of these sets have degrees summing to at most $p_1$ and those acting faithfully on the second set have degrees summing to at most $p_2$, and since each $G_{i}$ must act faithfully on at least one of the two sets, we get $n \le p$, contrary to assumption.

So $G$ is transitive. Suppose next that it is imprimitive, with blocks $p_2$ blocks of size $p_1$, where $p_1,p_2>1$ and $p_1p_2=p$. Some of the $G_{i}$ act faithfully on the set of $p_2$ blocks and, by induction, there degrees sum to at most $p_2$. The remaining $G_i$ either lie in the kernel of the action on the block system, or else $n_i>2$ and $A_{n_i}$ lies in the kernel of this action. Since the actions of the kernel on each of the blocks are equivalent, they must all act faithfully on each individual block of size $p_1$. So, by induction, the sum of their degrees is at most $p_1$ and hence $n \le p_1 + p_2$ which is not possible, since $n > p = p_1p_2$.

So $G$ must be primitive of degree $p$. Since normal subgroups of primitive groups are transitive, $S_{n_1}$ is transitive, and hence $p \le n_1!$. Now, the centralizer of a transitive group of degree $p$ has order at most $p$ (its order is equal to the number of fixed points of the point stabilizer of the transitive group being centralized), so we get $n_2!n_3! \cdots n_k! \le p$, which is impossible except when $k=2$ and $p=n_1!=n_2!$, but then $n_1+n_2 \le p$ (except when $n_1=n_2=2$), contrary to assumption.