(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.

(Edited after Christian's comments.)

For $0\le i\le n^{4/7}$, $$n!/(n-i)! = n^i \exp(-i(i-1)/(2n)+O(i^3/n^2)).$$ 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 $e^{-\Omega(n^{1/7})}$.

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

Using the fact that the $(n-i)! = n^i \exp(-i(i-1)/(2n)+O(i^3/n^2))$ and replacing the sum by an integral (an easy gaussian) we find that the sum is $$ \sqrt{\frac{\pi n}{2}} + O(1). $$ Confirmed numerically.

(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.