The answer is $\frac{n-1}{2}$ if $n$ is odd, $\frac{n+2}{2}$ if $n \ge 6$ is even, and $4$ if $n = 4$.

Let $A$, $B$, $C$ be the sets of friends of $A$, $B$, $C$ respectively excluding $A$, $B$, $C$ themselves.

Case 1. $n$ is odd

Since each of $A$, $B$, $C$ has at least $\frac{n+1}{2}$ friends, we have $|A|, |B|, |C| \ge \frac{n-3}{2}$. For any person $X$ other than $A$, $B$, $C$, the set $\{A, B, X\}$ is a friendly group if and only if $X \in (A \cap B)$. Similarly, $|B \cap C|$ and $|C \cap A|$ count the number of friendly groups of the form $\{B, C, X\}$ and $\{C, A, X\}$ respectively. Together with the friendly group $\{A, B, C\}$, by the inclusion-exclusion principle, the number of friendly groups is at least

$$|A \cap B| + |B \cap C| + |C \cap A| + 1 = |A| + |B| + |C| + |A \cap B \cap C| - |A \cup B \cup C| + 1 \ge \frac{n-3}{2} \times 3 + 0 - (n-3) + 1 = \frac{n-1}{2}.$$

The minimum value $\frac{n-1}{2}$ can be attained. For example, suppose $\frac{n-3}{2}$ people are friends of $A$ and $B$, and the other $\frac{n-3}{2}$ people are friends of $C$. Then $\{A, B, C\}$ is a friendly group and all other $\frac{n-3}{2}$ friendly groups are of the form $\{A, B, X\}$.

Case 2. $n$ is even and $n \ge 6$

Similarly, we have $|A|, |B|, |C| \ge \frac{n-2}{2}$. The number of friendly groups is at least
$$|A \cap B| + |B \cap C| + |C \cap A| + 1 = |A| + |B| + |C| + |A \cap B \cap C| - |A \cup B \cup C| + 1 \ge \frac{n-2}{2} \times 3 + 0 - (n-3) + 1 = \frac{n+2}{2}.$$

The minimum value $\frac{n+2}{2}$ can be attained. For example, suppose $\frac{n-4}{2}$ people are friends of $A$ and $B$, one person is a friend of $B$ and $C$, one person is a friend of $C$ and $A$, and the other $\frac{n-6}{2}$ people are friends of $C$. Then $\{A, B, C\}$ is a friendly group. In addition, there are $\frac{n-4}{2}$ friendly groups of the form $\{A, B, X\}$, one friendly group of the form $\{B, C, Y\}$, and one friendly group of the form $\{C, A, Z\}$.

Case 3. $n = 4$

Since each of $A$, $B$, $C$ has more than $2$ friends, every pair of the four people must be friends of each other. Clearly, there are $4$ friendly groups.