First, estimate the upper bound of $f(n)$, i.e., prove the right inequality in (2).
Each three-element subset $\{i, j, k\}$ contains three two-element subsets $\{i, j\}$, $\{j, k\}$, $\{i, k\}$.

Since every two elements (three-element subsets of $X$) in $\mathscr{A}$ have at most one common element, the two-element subsets contained in each pair of three-element subsets are all different.
The two-element subsets of $X$ total $C_{n}^{2}$, so
$$3 f(n) \leqslant C_{n}^{2},$$

i.e.,
$$f(n) \leqslant \frac{1}{3} C_{n}^{2}=\frac{n^{2}-n}{6} .$$

To estimate the lower bound of $f(n)$, the construction method should be used. Construct a batch of three-element subsets, the number of which is $\geqslant \frac{1}{6} n(n-4)$, and each pair of them has at most one common element.
For this purpose, consider all three-element subsets $\{i, j, k\}$ that satisfy the condition
$$i+j+k \equiv 0 \quad(\bmod n)$$
(i.e., $i+j+k$ is divisible by $n$).
If $i^{\prime}=i \cdot j^{\prime}=j$, and
$$i^{\prime}+j^{\prime}+k^{\prime} \cong i+j+k \equiv 0 \quad(\bmod n),$$

then
$$k^{\prime} \equiv k \quad(\bmod n) .$$

When $k^{\prime}, k \in\{1,2, \cdots, n\}$, (4) is $k^{\prime}=k$. Therefore, every two (different) three-element subsets that satisfy (3) have at most one common element.
Now, calculate the number $s$ of three-element subsets $\{i, j, k\}$ that satisfy (3).
First, choose $i$, which has $n$ choices. After $i$ is chosen, choose $j, j \neq i$, and $j$ does not satisfy the congruence equation
$$2 i+j=0 \quad(\bmod n)$$
(i.e., when $2 i<n$, $j \neq n-2 i$; when $2 i \geqslant n$, $j \neq 2 n-2 i$) and
$$i+2 j \equiv 0 \quad(\bmod n)$$
(i.e., $\left.j \neq \frac{n-i}{2}, j \neq \frac{2 n-i}{2}\right)$. Therefore, $j$ has at least $n-4$ choices. After $i$ and $j$ are determined, $k$ is also determined by (3), and is different from $i$ and $j$. Therefore, $s \geqslant \frac{1}{6} n(n-4)$. Thus, the other half of (2) holds.