Since in each round play exactly 3 players, the number of couples playing at each round is $\binom{3}{2} = 3$. Therefore, at the end of the game after $n$ rounds, the total number of couples played the game will be $3n$. According to the last rule:
$$\binom{n}{2} \le 3n \Leftrightarrow \frac{n(n-1)}{2} \le 3n \Leftrightarrow \frac{n-1}{2} \le 3 \Leftrightarrow n \le 7.$$
Next we will prove that the value $n=7$ is possible. In fact, for $n=7$ we have $\binom{7}{2} = \frac{7!}{2!5!} = \frac{6 \cdot 7}{2} = 21 = 3 \cdot 7$. If the friends are: $A$, $B$, $\Gamma$, $\Delta$, $E$, $Z$, $H$, then we can define the following triads $(A, B, \Gamma)$, $(A, \Delta, E)$, $(A, Z, H)$, $(B, \Delta, H)$, $(B, E, Z)$, $(\Gamma, \Delta, Z)$, $(\Gamma, E, H)$.