For $n$ dice with $3$ faces numbered $1$ to $n$, $n+1$ to $2n$, and $2n+1$ to $3n$, the numbers range from $1$ to $3n$.

The total sum of all of these is $${3n(3n+1)\over 2}.$$

We must divide this among the $n$ dice, so each dice should add up to $${3(3n+1)\over 2}.$$

For $5$ dice, this would be $3(15+1)/2$ or $24$.