Here's 14 sets on 8 points:

{ 2, 5, 7, 8 }, { 2, 3, 6, 7 }, { 2, 3, 4, 5 }, { 2, 4, 6, 8 }, { 1, 2, 4, 7 }, { 1, 2, 3, 8 }, { 3, 5, 6, 8 }, { 1, 2, 5, 6 }, { 1, 3, 5, 7 }, { 1, 4, 5, 8 }, { 3, 4, 7, 8 }, { 1, 6, 7, 8 }, { 1, 3, 4, 6 }, { 4, 5, 6, 7 }

It is a clique problem in the fusion of 2 classes of an association scheme, so we might be able to get bounds. But given that we don't even know the maximum size of cliques in Johnson graphs, it may be difficult to give the exact answer.

But its not $\binom{n/2}{2}$.

Here's 14 sets on 8 points:

{ 2, 5, 7, 8 }, { 2, 3, 6, 7 }, { 2, 3, 4, 5 }, { 2, 4, 6, 8 }, { 1, 2, 4, 7 }, { 1, 2, 3, 8 }, { 3, 5, 6, 8 }, { 1, 2, 5, 6 }, { 1, 3, 5, 7 }, { 1, 4, 5, 8 }, { 3, 4, 7, 8 }, { 1, 6, 7, 8 }, { 1, 3, 4, 6 }, { 4, 5, 6, 7 }

It is a clique problem in the fusion of 2 classes of an association scheme, so we might be able to get bounds. But given that we don't even know the maximum size of cliques in Johnson graphs, it may be difficult to give the exact answer.

But its not $\binom{n/2}{2}$.

Actually, I retract that. It might be $\binom{n/2}{2}$, but not for small $n$.