Let $a=\cos A + i\sin A, b=\cos B + i\sin B,$ and $c=\cos C + i\sin C$. Then \begin{align} &\cos 3A + \cos 3B + \cos 3C = 1\\ \Leftrightarrow\quad & a^3 + \frac{1}{a^3}+b^3 + \frac{1}{b^3} + c^3 +\frac{1}{c^3} = 2\\ \Leftrightarrow\quad & a^3 - {b^3c^3} + b^3-{c^3a^3} + c^3-{a^3b^3} = 2\\ \Leftrightarrow\quad&a^3+b^3+c^3-a^3b^3-b^3c^3-c^3a^3=1-a^3b^3c^3\\ \Leftrightarrow\quad&(a^3-1)(b^3-1)(c^3-1)=0 \end{align} This implies one of $a, b,$ and $c$ must be $\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$. Therefore we conclude that the largest angle must be $\boxed{\frac{2\pi}{3}}$ or $\boxed{120^\circ}$.