I would guess that the answer is $O(n!/n)$. You can get an example of this size as follows. Partition $\Omega = \{1, \dots, n\}$ as $\{1\} \cup T \cup S$ where $|T| = t$, and let $X$ be the set of all $\pi$ such that $\pi(1), \pi^{-1}(1) \in T$ and $\pi(T) \subset S$. The density of $X$ is then comparable to $(t/n)^2 (1 - t/n)^t \approx (t/n)^2 \exp(-t^2/n)$. The best choice for $t$ is $cn^{1/2}$ for some constant $c$. Meanwhile, by design, $X^{-1} \cap XX = \emptyset$, so $XXX$ does not contain $1$.

Can anybody spot a better construction?

More is known about a slight variant of your question. Suppose you want to know the minimal density $\alpha$ such that if $X, Y, Z$ each have density at least $\alpha$ then $XYZ = G$. Let $G = A_n$. Gowers's method gives the upper bound $n^{-1/3}$. An example like the one above gives a lower bound $n^{-1/2}$. The truth is $n^{-1/2+o(1)}$. I wrote a paper about this a few years ago, which you can find here: https://arxiv.org/abs/1512.03517.