It seems to be the case that the polynomial $P_n(x) =\sum_{m=1}^n m!S(n,m)x^m$ has only real zeros. (I know it is true that $\sum_{m=1}^n S(n,m)x^m$ has only real zeros.) If this is true, then the value of $m$ maximizing $m!S(n,m)$ is within 1 of $P'_n(1)/P_n(1)$ by a theorem of J. N. Darroch, Ann. Math. Stat. 35 (1964), 1317-1321. See also J. Pitman, J. Combinatorial Theory, Ser. A 77 (1997), 279-303. By standard combinatorics $$ \sum_{n\geq 0} P_n(x) \frac{t^n}{n!} = \frac{1}{1-x(e^t-1)}. $$ Hence $$ \sum_{n\geq 0} P_n(1)\frac{t^n}{n!} = \frac{1}{2-e^t} $$ $$ \sum_{n\geq 0} P'_n(1)\frac{t^n}{n!} = \frac{e^t-1}{(2-e^t)^2}. $$ Since these functions are meromorphic with smallest singularity at $t=\log 2$, it is routine to work out the asymptotics, though I have not bothered to do this.
Followup
Update. It is indeeed indeed true that $P_n(x)$ has real zeros. This is because $(x-1)^nP_n(1/(x-1))=A_n(x)/x$, where $A_n(x)$ is an Eulerian polynomial. It is known that $A_n(x)$ has only real zeros, and the operation $P_n(x) \rightarrow to (x-1)^nP_n(1/(x-1))$ leaves invariant the property of having real zeros.
