By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of side length $m$, the number of regular hexagons whose enveloping lattice hexagon is $G$ is exactly $m$.

Yet also there are precisely $3(n-m)(n-m+1)+1$ lattice hexagons of side length $m$ in our lattice: they are those with centres lying at most $n-m$ steps from the centre of the lattice. In particular, the total number of regular hexagons equals
$$N = \sum_{m=1}^{n} (3(n-m)(n-m+1)+1)m = (3n^2+3n) \sum_{m=1}^{n} m - 3(2m+1) \sum_{m=1}^{n} m^2 + 3 \sum_{m=1}^{n} m^3.$$
Since $\sum_{m=1}^{n} m = \frac{n(n+1)}{2}$, $\sum_{m=1}^{n} m^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{m=1}^{n} m^3 = \left(\frac{n(n+1)}{2}\right)^2$ it is easily checked that $N = \left(\frac{n(n+1)}{2}\right)^2$.