I claim $a_i=n$, $1\le i\le n$, is the only solution. Note that using Cauchy-Schwarz, $\textstyle \sum_i a_i^2 \ge \frac1n(\sum_i a_i)^2 \ge n^3$. Consequently, $a_1^2+\cdots+a_n^2\in\{n^3,n^3+1\}$. Assume there is such an $(a_1,\dots,a_n)$ such that $a_1^2+\cdots+a_n^2=n^3+1$. Then using the fact $\textstyle \sum_i a_i^2\equiv \sum_i a_i\pmod{2}$, it follows that $\textstyle\sum_i a_i\ge n^2+1$. Applying Cauchy-Schwarz once again, we find $n(n^3+1)\ge (n^2+1)^2\iff n\ge 2n^2+1$, which is absurd. Hence $\textstyle\sum_i a_i^2=n^3$ and $\textstyle \sum_i a_i = n^2$. From here, we have the equality. Using the fact \[ n\sum_i a_i^2 - \left(\sum_i a_i\right)^2 = \sum_{i,j}(a_i-a_j)^2=0, \]we immediately get $a_i=n$ for all $i$.