Consider the cases:
1. Use 1 color: $\binom{n}{1} = n$,

2. Use 2 colors: $4 \times \binom{n}{2} = 2n(n-1)$,

3. Use 3 colors: $9 \times \binom{n}{3} = 3 \frac{n(n-1)(n-2)}{2}$,

4. Use 4 colors: $6 \times \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{4}$

The sum up will give:
$$\frac{n^{4} + n^{2} + 2n}{4}$$

Remark. We can rotate the coloring with respect to the angles $0, \frac{2\pi}{4}, 2\frac{2\pi}{4}, 3\frac{2\pi}{4}$. We get $4 \times$ (the number $S$ of the rotationally distinct colorings). The arbitrary rotation of the colorings with only 1 color gives the same coloring, thus we have to add $3 \times \binom{n}{1}$ and similarly $2 \times \binom{n}{2}$. Thus, we get
$$4 \times S = n^{4} + 3 \times \binom{n}{1} + 2 \times \binom{n}{2} = n^{4} + n^{2} + 2n \Rightarrow S = \frac{n^{4} + n^{2} + 2n}{4}.$$