The 12 friends will play $\binom{12}{2} = \frac{12 \cdot 11}{2} = 66$ games, so the total number of points from all games is $66$. One possible outcome of the tournament is to put all players in an order and each of them to win all players that are after of him in the order. Then the first one has $11$ points, the second one $10$ points and so on. Then the last one has $0$ points, and all the players have a different number of points from $0$ to $11$. Then:
$$\Sigma_3 = 0^3 + 1^3 + \dots + 11^3 = \left(\frac{11 \cdot 12}{2}\right)^2 = 66^2$$
We will prove that this is the maximum value for $\Sigma_3$.
Indeed, suppose that there is another outcome of the tournament which gives maximum $\Sigma_3$ and in that tournament the players $A$ and $B$ have the same number of points, let it be $\kappa$, where $A$ won $B$. We observe that $\kappa \neq 0$ since $A$ won against $B$. A new outcome will be produced if we suppose that $B$ won against $A$, so $A$ would have $\kappa - 1$ points and $B$ will have $\kappa + 1$ points. If all the other games had the same outcome, then the difference between the new $\Sigma_3$ and the old one will be
$$(\kappa + 1)^3 + (\kappa - 1)^3 - 2\kappa^3 = 6\kappa > 0,$$
so the new $\Sigma_3$ is greater than the old one which we supposed that has the maximum $\Sigma_3$, a contradiction.
It follows that the maximum $\Sigma_3$ equals to $66^2$ and is reached with the above example.