Solution:

Denote by $A$, $B$ and $C$ the three familiar people in the group.
Let $n=2k+1$ be an odd integer. Then any of $A$, $B$ and $C$ has at least $k+1$ familiar ( $k-1$ of them are not $A$, $B$ or $C$ ). Denote by $T$ the set of all people except $A$, $B$ and $C$ and let $a_{i}$, $i=0,1,2,3$, be the set of the people in $T$ who have exactly $i$ familiar among $A$, $B$ and $C$.
Then $a_{0}+a_{1}+a_{2}+a_{3}$ is the number of all members of $T$, i.e. we have
$$a_{0}+a_{1}+a_{2}+a_{3}=2k-2$$
On the other hand, $a_{1}+2a_{2}+3a_{3}$ is the number of all familiar to $A$, $B$ and $C$, i.e. we have
$$a_{1}+2a_{2}+3a_{3} \geq 3k-3$$
Hence
$$\begin{aligned} 3k-3 &\leq a_{1}+2a_{2}+3a_{3}=a_{0}+a_{1}+a_{2}+a_{3}+a_{2}+2a_{3} \\ &=2k-2+a_{2}+2a_{3} \end{aligned}$$
and therefore $a_{2}+2a_{3} \geq k-1$.
Since any familiar to two of $A$, $B$ and $C$ is a member of a triple of familiar people and any familiar to $A$, $B$ and $C$ is member of three such triples, then the number of these triples is at least $1+a_{2}+3a_{3}$. Thus $1+a_{2}+3a_{3}>a_{2}+2a_{3} \geq k-1$, which means that the number of the triples is not less than $k$.
It remains to construct an example with $k$ triples of familiar people. Let there be no familiar people in $T$. If $A$ is familiar to exactly $k-1$ people of $T$, and $B$ and $C$ to the remaining $k-1$, then the number of the triples is $k$.

Let $n=2k$ be even. As in the previous case, we get that the number of the triples of familiar people is at least $k+1$. If $A$ and $B$ have exactly one common familiar person from $T$ (it is possible, since $|T|=2k-3$ and the familiar to $A$ and $B$ are at least $k-1$ ) who is not familiar to $C$, then the number of the triples is exactly $k+1$.

So the answer of the problem is $k$ for $n=2k+1$ and $k+1$ for $n=2k$.