It is easy to be seen that the left hand side is divisible by $9$ implying that the equation has no solution for $n = 1$.

Let $n=2$. We have $2013^2 = (3 \cdot 11 \cdot 61)^2$, $3^2 = 2^3 + 1^3$ and $11 \cdot 61 = 8^3 + 4^3 + 1^3 + 3 \cdot 8 \cdot 4 \cdot 1$. We use the identity
$$(a^3 + b^3 + c^3 - 3abc)(x^3 + y^3 + z^3 - 3xyz) = u^3 + v^3 + w^3 - 3uvw.$$
where $u = ax + by + cz$, $v = ay + bz + cx$ and $w = az + bx + cy$. We apply the above identity first for $(8, 4, -1)$ and $(4, 8, -1)$, and then for the obtained triples $(65, 0, 56)$ and $(2, 1, 0)$ and obtain $65^3 + 56^3 = (11 \cdot 61)^2$ and $130^3 + 177^3 + 56^3 - 3 \cdot 130 \cdot 177 \cdot 56 = 2013^2$.