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The answer is $\Theta (n^2)$. Partition all but at most one point into $m=\lfloor n/2\rfloor$ disjoint pairs and take all unions of two pairs. There are $\left(\genfrac{}{}{0ex}{}{m}{2}\right)=\Theta (n^2)$ such four-sets, and two distinct unions intersect in zero or two points. The Ray-Chaudhuri–Wilson theorem gives the upper bound $O(n^2)$.

The answer is $\Theta (n^2)$. All four-sets containing one fixed pair form a family of sizewhose distinct intersections have size two or three. The Ray-Chaudhuri–Wilson theorem for the two allowed intersection sizes gives the matching $O(n^2)$ upper bound.

The answer is $\Theta (n^2)$. Take a Steiner triple system, or a partial one of quadratic size, on $\{2,\dots ,n\}$ and adjoin the fixed point $1$ to every triple. Distinct triples meet in zero or one point, so the resulting four-sets meet in one or two points. This gives $\Omega (n^2)$ members. The Ray-Chaudhuri–Wilson theorem with the two allowed intersection sizes gives $O(n^2)$.