For any positive integer $n$, write $n!=a_nb_n^2$, where $a_n$ is squarefree. Does there exist a constant $c$ such that for any $\epsilon>0$, there exists $N$ such that $c^{(1-\epsilon)n}<a_n<c^{(1+\epsilon)n}$ for all $n>N$?
From Stirling's approximation we know that $n!$ is on the order of $\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$. The product of all primes less than $n$ is roughly $e^n$, as shown by Erdos. The product we are considering can only be less than that.
Note: the question first indicated $2^n$ as an estimate, which is correct. Edited since.
ORIGINAL: Not a bad guess. I get, with Chebyshev's first function $\theta(x),$ the log of your number as $$ \theta(n) - \theta \left( \frac{n}{2} \right) +\theta \left( \frac{n}{3} \right) -\theta \left( \frac{n}{4} \right) +\theta \left( \frac{n}{5} \right) -\theta \left( \frac{n}{6} \right) \cdots $$ where the alternating series $1 - \frac{1}{2} +$ has limit $\log 2.$
The main ingredient is Legendre's theorem on the power of a prime dividing a factorial. Much more work to establish explicit bounds, if true.
Note that oeis agrees https://oeis.org/A055204 but with no proof or error estimates.