Estimate related to the Möbius function

You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.

I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula $$\sum_\text{$n$ square-free}z^{\omega(n)}n^{-s}=\prod_p\left(1+\frac{z}{p^s}\right).$$ Then you will need to "factor out" $\zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.

It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas \begin{align*} \#\{ n\le x\colon \omega(n) = k \} &\sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!} \\ \#\{ n\le x\colon \Omega(n) = k \} &\sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!}. \end{align*} (Here $\omega$ and $\Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)

Note that any nonsquarefree integer $n$ with $\Omega(n) = k$ satisfies $\omega(n) = j$ for some $1\le j\le k-1$. Therefore the bounds \begin{align*} \#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} &\le \#\{ n\le x\colon \Omega(n) = k \} \\ \#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} &\ge \#\{ n\le x\colon \Omega(n) = k \} - \sum_{j=1}^{k-1} \#\{ n\le x\colon \omega(n) = j \}, \end{align*} together with the Selberg–Sathe asymptotic formulas above, immediately imply that $$ \#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} \sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!}. $$

This argument can be made to work with some uniformity in $k$ as well (I think $k=o(\log\log x)$ is what is needed).