A formula as a single sum is
$$f_{n,k} = \binom nk \sum_{i=0}^k \left(-\frac12\right)^i (k+i)\,i!\, \binom{k}{i}\binom{n-k}{i} n^{n-k-i-1}.$$
This formula can be found in J. W. Moon's *Counting Labelled Trees*, Theorem 4.1. He attributes it to A. Rényi, *Some remarks on the theory of trees*, Publications of the Mathematical Institute of the Hungarian Academy of Sciences 4 (1959), 73--85.

To prove it, let $T= \sum_{n=1}^\infty n^{n-1} x^n/n!$ be the exponential generating function for rooted trees and let $U=\sum_{n=1}^\infty n^{n-2} x^n/n!$ be the exponential generating function for unrooted trees. It is well known that
$$T^k = k! \sum_{n=k}^\infty \binom nk k n^{n-k-1}\frac{x^n}{n!}.\tag{$*$}$$
Then $U = T-T^2/2$ so the exponential generating for forests of $k$ trees is
$$\sum_{n=k}^\infty f_{n,k} \frac{x^n}{n!} = \frac{U^n}{n!}=\frac{(T-T^2/2)^k}{k!},$$
and the formula for $f_{n,k}$ follows by expanding the right side by the binomial theorem and using $(*)$.