How many numbers below $n$ with a square number of distinct prime factors?

Let $\pi_k(n)$ be the number of integers $\le n$ with exactly $k$ distinct prime factors. You are looking for $\sum_{m=1}^{\infty}\pi_{m^2}(n)=\pi(n)+\pi_4(n)+\pi_9(n)+\ldots$. Hardy and Ramanujan proved the upper bound $$ \pi_k(n) < c(n/\log n) \cdot (\log\log n + d)^{k-1}/(k-1)!$$ for some constants $c$ and $d$, and Erdös and Pillai (independently) proved the lower bound $$ \pi_k(n) > c'(n/\log n) \cdot (\log\log n)^{k-1}/(k-1)! $$ for some constant $c'$. So, putting things together, you have $$ \frac{c'(\log\log n)^{k-1}}{(k-1)!}<\frac{\pi_{k}(n)}{n/\log n}<\frac{c(\log\log n+d)^{k-1}}{(k-1)!}. $$ Taking $g(z)=\sum_{m=1}^{\infty}\frac{z^{m^2-1}}{(m^2-1)!}$, you have $$ c'g(\log \log n)<\frac{\sum_{m=1}^{\infty}\pi_{m^2}(n)}{n/\log n}<cg(\log\log n + d). $$ Update:

After spending some time looking into the asymptotic growth of functions like $g(z)$, analytic with every derivative at the origin equalling zero or one... which seems interesting and may be worth its own question... I think that $g(z)$ is $\Theta(e^z / \sqrt{z})$ for large $z$. If that's correct, then in fact $$ \sum_{m=1}^{\infty}\pi_{m^2}(n)\in\Theta\left(\frac{n}{\sqrt{\log\log n}}\right). $$