any ideas?

The case $d=1$ is achievable at $(a_{1},a_{2},...,a_{n})=(2,1,1,...,1),$ so let $d>1$. Let $P=a_{1}a_{2}\cdots a_{n},$ notice that $$a_{1}^{n}\equiv a_{2}^{n}\equiv ...\equiv a_{n}^{n}\equiv -P\;(mod\;d)$$so $\gcd (d,P)=1$ (otherwise $p|\gcd (a_{1},a_{2},...,a_{n})$ for some prime divisor $p$ of $d$). Now $$P^{n}\equiv \prod _{k=1}^{n}a_{k}^{n}\equiv (-P)^{n}\equiv -P^{n}\;(mod\; d)$$Which immediately gives $d=2$, and it is achievable at $a_{1}=a_{2}=...=a_{n}$. So we are done

The same as 2017 ELMO.