Let's turn the question around and let $f(n)$ be the number of consecutive integers required to guarantee that $n$ of them are pairwise relatively prime. E.g., $f(4)=6$ because you can find a set of 5 consecutive integers no 4 of which are pairwise relatively prime ($\lbrace2,3,4,5,6\rbrace$ will do) but given any set of 6 consecutive integers there must be 4 that are relatively prime (the three odd ones are pairwise relatively prime, and there will be an even that's not a multiple of 3 or 5, and it will be relatively prime to each of the odds).
I think that $f(n)=h(n-1)$, where $h(n)$ is (based on) the Jacobsthal function: $h(n)$ is the number of consecutive integers required to guarantee that one of them will not be a multiple of any of the first $n$ primes. E.g., $h(3)=6$ because you can find a set of 5 consecutive integers each of which is divisible by (at least) one of 2, 3, or 5 ($\lbrace2,3,4,5,6\rbrace$ will do) but given 6 consecutive integers only 3 can be even, and of the three odds, only one can be a multiple of 3, and only one can be a multiple of 5.
Now, why should $f(n)=h(n-1)$? Well, if you have $n$ pairwise coprime integers, it must be the case that (at least) one is not divisible by any of the first $n-1$ primes, for if each of your $n$ integers is divisible by one (or more) of the first $n-1$ primes, then two of them must be divisible by the same prime, hence, not relatively prime. Thus, $f(n)\ge h(n-1)$. I can't quite see my way through a proof that $h(n)\ge f(n+1)$, so there's still some work to be done here.
Anyway, the point is, lots of work has been done on the Jacobsthal function, including estimates. A good reference is Thomas R Hagedorn, Computation of Jacobsthal's function $h(n)$ for $n\lt50$, Math Comp 78 (2009) 1073-1087. As the title indicates, the paper is mostly concerned with computing Jacobsthal's function, but the author does summarize the history and gives many references.
Forgetting the squarefree condition for a moment, the number of integers up to $N$ that are not divisible by any primes less than $N^{1/k}$ is asymptotic to
$$
\omega(k) \frac N{\log N} \sim e^\gamma \omega(k) N \prod_{p\le N^{1/k}} \bigg( 1-\frac1p \bigg),
$$
where $\omega$ is the Buchstab function. (In particular, the first half of the density you derive in your answer is not correct: there is a correction factor of the form $e^\gamma \omega(k)$.)
Now the number of integers up to $N$ that are divisible by the square of a particular prime $p$ is at most $N/p^2$. So the number of integers up to $N$ that are divisible by the square of a prime greater than $N^{1/k}$ is at most
$$
\sum_{p>N^{1/k}} \frac N{p^2} < N \sum_{n>N^{1/k}} \frac1{n^2} < N \cdot \frac1{N^{1/k}}.
$$
Therefore the above asymptotic also holds for squarefree numbers not divisible by small primes.
Best Answer
By partial summation $$ s(n) = n\pi(n)-\sum_{m=2}^{n-1}\pi(m) $$ so by the Prime Number Theorem $$ s(n) = \frac{n^2}{\log n}-\sum_{m=2}^{n-1}\frac{m}{\log m}+O\left(\frac{n^2}{\log^2 n}\right). $$ The sum on the right is $$ \sum_{m=2}^{n-1}\frac{m}{\log m} = \int_2^n \frac{x}{\log x}dx + O\left(\frac{n}{\log n}\right) $$ using the monotonicity properties of the integrand. Now the integral equals, by partial integration, $$ \int_2^n \frac{x}{\log x}dx = \left[\frac{x^2}{2\log x}\right]_2^n + \int_2^n \frac{x}{2\log^2 x}dx = \frac{n^2}{2\log n} + O\left(\frac{n^2}{\log^2 n}\right).$$ Altogether we have $$ s(n) = \frac{n^2}{2\log n} + O\left(\frac{n^2}{\log^2 n}\right).$$ This can be made more precise both numerically and theoretically.