To complete the method sketched in the comments.
The answer is $\binom {13}5$.
To see this, suppose we are choosing $5$ values from $\{1,\cdots, 13\}$. Order the choices as $n_1<\cdots <n_5$. Let $n_3=k+1$. Clearly $2≤k≤10$. Given a choice of such an $n_3$ we see that we are asked to choose $2$ values from $\{1,\cdots, k\}$ and $2$ from $\{k+2,\cdots, 13\}$. There are $\binom k2$ ways to do the former and $\binom {13-(k+2)+1}2=\binom {12-k}2$ ways to do the latter. Hence the number of choices with a given $n_3$ is the product $\binom {k}2\times \binom {12-k}2$. Summing over $k$ we see that $$\binom {13}5=\sum_{k=2}^{10}\binom {k}2\times \binom {12-k}2$$ as desired.