Since $346^{346}\equiv 4 \pmod{9}$, we need at least four perfect cubes. On the other hand, $(7\cdot346^{115})^3 + (346^{115})^3 + (346^{115})^3 + (346^{115})^3 = (7^3+1^3+1^3+1^3)\cdot346^{345}=346^{346}$. So the minimum number is $\boxed{4}$.