It is a consequence of the following algebraic identity
$1+2^{2}+3^{2}+\ldots +n^{2}=\frac{1}{3}\left( n^{3}+3n^{2}+3n+1\right) - \frac{1}{3}n-\frac{1}{2}(n^2+n)-\frac{1}{3}.$ $\tag{1}$
The RHS is a cubic function of $n$: $\frac{1}{3}n^{3}+\frac{1}{2}n^{2}+\frac{1}{6}n$.
Proof. From the algebraic identity
$\left( 1+k\right) ^{3}=1+3k+3k^{2}+k^{3}$
we get the following $n$ identities:
$\begin{eqnarray*} \left( 1+1\right) ^{3} &=&1+3+3+1 \\ \left( 1+2\right) ^{3} &=&1+3\cdot 2+3\cdot 2^{2}+2^{3} \\ \left( 1+3\right) ^{3} &=&1+3\cdot 3+3\cdot 3^{2}+3^{3} \\ &&\ldots \\ \left( 1+n\right) ^{3} &=&1+3n+3n^{2}+n^{3}. \end{eqnarray*}$
Now if we sum these equalities and cancel the common terms, $\left( 1+1\right) ^{3}$ on the LHS of the first and $2^{3}$ on the RHS of the second, $\left( 1+2\right) ^{3}$ on the LHS of the second and $3^{3}$ on the RHS of the third, etc., and $(1+n-1)^3$ on the LHS of the second last and $n^3$ on the RHS of the last, we get:
$\left( 1+n\right) ^{3}=n+3(1+2+3+\ldots +n)+3(1+2^{2}+3^{2}+\ldots +n^{2})+1,\tag{2}$
and $(1)$ follows from $(2)$ and the sum formula for the arithmetic progression $1+2+3+\ldots +n=\frac{\left( n+1\right) n}{2}.$
(Adapted from Proof 1 in this answer to the question Prove that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$? ; see the above Theo Buehler's comment.)
Note: This is a particular case of the sum $1+2^{p}+3^{p}+\ldots +n^{p}$, which is a polynomial of degree $p+1$.