First, trivial to check that $p_1=\dotsm =p_{13}=2$ fails. So $p_{13}>2$. Now consider the equation modulo $8$. Suppose $n$ of the $p_i$'s on the LHS are $2$ and $12-n$ are not. Then $4n+1(12-n)\equiv 1\pmod{8}\implies 3n\equiv 5\pmod{8}\implies n=7$, since $n<12$. Now we have $7\cdot 4+p_8^2+\dots +p_{12}^2=p_{13}^2$. Consider this modulo $3$. Let $k$ of the remaining primes on the LHS be $3$, then the equation becomes $28+5-k\equiv 1\pmod{3}$ (since $p_{13}>3$) and therefore $k\equiv 2\pmod{3}$, so $k=2,5$. But clearly $k=5$ fails, since then $7\cdot 2^2+5\cdot 3^2=73$, not a square. So $k=2$, and $28+18+p_{10}^2+p_{11}^2+p_{12}^2=p_{13}^2$. Consider this modulo $5$. Then $p_{10}^2+p_{11}^2+p_{12}^2\equiv 0\pmod{5}$, since $p_{13}>5$. Now this implies that either all three are $5$, or exactly one is $5$. Clearly if all three are $5$, then $p_{13}=11$, but none are equal to $2p_1+p_9=7$. Therefore $p_{10}=5$ and we obtain $71+p_{11}^2+p_{12}^2=p_{13}^2$. Since $2p_1+p_9=7$, at least one of $p_{11}$, $p_{12}$ is $7$, so $p_{11}=7$. If $p_{12}=7$, then $p_{13}=13$. If $p_{12}>7$, then we have possible pairs $(p_{12},p_{13})=(13,17),(29,31)$. In summary, the solutions areas follows: We must have $a_1=\dots =a_7=2$, $a_8=a_9=3$, $a_{10}=5$, $a_{11}=7$. Then we have the possible pairs $(a_{12},a_{13})=(7,13),(13,17),(29,31)$.