@Haran enquired about case for $n=6$.

Well lately Seiji Tomita has given a numerical solution for it. 

He used the equation below to arrive at a numerical solution:
$$
(m^2+m+1)^3+(m^2-m+1)^3+(2)^3=(m^2+5)(m^2+m+1)(2)(m^2-m+1).
$$

To arrive at $n=6$ we substitute $m=1$. The downside of this is that, when using the above equation one can get only one numerical solution for $n=6$. The equation gives us:
$$
3^3+1^3+2^3=6(3\cdot2\cdot1).
$$
Tomita has given a general method through which solutions for different $n$ can be arrived at. His ["Computational number theory" web page](http://www.maroon.dti.ne.jp/fermat/eindex.html) discusses this in [#456. $x^3 + y^3 + z^3 = n x y z$](http://www.maroon.dti.ne.jp/fermat/dioph448e.html).