We say a finite poset $P$ is a tree poset if its Hasse diagram is a tree. Let $k$ be the length of the largest chain contained in $P$. We show that when $P$ is a fixed tree poset, the number of $P$-free set systems in $2^{[n]}$ is $2^{(1+o(1))(k-1){n \choose \lfloor n/2\rfloor}}$. The proof uses a generalization of a theorem by Boris Bukh together with a variation of the multiphase graph container algorithm.