I believe that this simplicial complex should be a simplex, therefore contractible. I should probably have known about Bertrand's postulate before this, but it says that for any $k$, there is a prime number $p$ with $k < p < 2k$. If $n$ is even, let $n=2k$ and find such a $p$. If $n$ is odd, say $n=2k-1$, then find $p$ with $k<p<2k$, and so we have $n/2 < k < p \leq n$.

Once you have such a prime, it is contained in every maximal coprime subset, and so the intersection of all of those sets will be nonempty: every vertex in the simplicial complex is contained in a single simplex.

Edit: this is answering the wrong question (coprime subsets rather than coprime-free subsets), but in case it's useful, I will leave it here.

I believe that this simplicial complex should be a simplex, therefore contractible. I should probably have known about Bertrand's postulate before this, but it says that for any $k$, there is a prime number $p$ with $k < p < 2k$. If $n$ is even, let $n=2k$ and find such a $p$. If $n$ is odd, say $n=2k-1$, then find $p$ with $k<p<2k$, and so we have $n/2 < k < p \leq n$.

Once you have such a prime, it is contained in every maximal coprime subset, and so the intersection of all of those sets will be nonempty: every vertex in the simplicial complex is contained in a single simplex.