I think that for large even $n=2k$, maybe $N:=\binom{k}{2}$ is the right answer.  

**Semi-proof.** Observe that $4N=2k(k-1)$.  Thus, if more than $N$ sets appear, then some element $x$ occurs in at least $k$ sets.  Removing $x$ from these $k$ sets, we get a family of $3$-subsets of a set of size $2k-1$ which pairwise intersect in $1$ element.  Now I think that for large $k$ this is not possible, although I am not an extremal set theorist.  I would guess that for large $k$ the maximum size of such a family is achieved by taking a family of disjoint $2$-sets and adding the same point to each set.  Such a family only has size $k-1$, which would be a contradiction.