We proved in this paper https://arxiv.org/abs/2012.01764 that the answer is $2^{n/4+o(n)}$. As observed in the other answers, a counting argument shows that this is optimal (up to the lower order term).
Update: May 7, 2021. Unfortunately there was a flaw in our proof and we have to take back this answer. The main result of the paper is still valid but we cannot obtain the universal poset as a direct consequence of our comparability labelling scheme for posets.
Very sorry, feel free to down-vote this answer !