(Edited after Christian's comment.)

For $0\le i\le n^{2/3}$,
$$n!/(n-i)! = n^i \exp(-i(i-1)/(2n)+O(i^3/n^2)).$$
Approximate the sum for that range by the corresponding integral (a gaussian with the right endpoint far into the tail). This can be formally justified by the Euler-Maclaurin theorem, and probably by elementary reasoning as the integrand is decreasing.

Bound the remainder of the sum by $n$ times the largest (first) term. This gives the negligible contribution $e^{-\Omega(n^{1/3})}$.

We find that the sum is
$$ \sqrt{\frac{\pi n}{2}} + O(1). $$
Confirmed numerically.