Analysis and Solution: From condition (1) $\left|A_{i}\right|=7(i=1,2, \cdots, n)$, we think of calculating the total number of times each element appears in all subsets $S_{1}$.

On the one hand, starting from each subset, we have $S_{1}=7 n$. On the other hand, starting from each element, let the number of times $i(i=1,2, \cdots, 15)$ appears be $r_{i}$, then we have $S_{1}=\sum_{i=1}^{15} r_{i}$, so $7 n=S_{1}=\sum_{i=1}^{15} r_{i}$. We only need to find the range of $\sum_{i=1}^{15} r_{i}$, a sufficient condition is to find the range of each $r_{i}$. Without loss of generality, we first find the range of $r_{1}$.

From condition (3), we think of calculating the number of all 3-element subsets containing 1, $S_{2}$.
On the one hand, starting from the whole $S$, we have $S_{2}=\mathrm{C}_{14}^{2}=91$. On the other hand, considering all $r_{1}$ subsets $A_{i}$ containing 1, each such subset contains $\mathrm{C}_{6}^{2}=15$ 3-element subsets containing 1, so there are a total of $15 r_{1}$ 3-element subsets containing 1. By condition (3), we know that such a calculation of 3-element subsets containing 1 does not miss any, so $15 r_{1} \geqslant S_{2}=91$, so $r_{1} \geqslant 7$. Similarly, for all $i=1,2, \cdots, 15$, we have $r_{i} \geqslant 7$. Thus, $7 n=\sum_{i=1}^{15} r_{i} \geqslant \sum_{i=1}^{15} 7=15 \times 7$, so $n \geqslant 15$.

When $n=15$, let $A_{i}=\{1+i-1,2+i-1,4+i-1,5+i-1,6+i-1, 11+i-1,13+i-1\}(i=1,2, \cdots, 15)$, if the number in the set is greater than 15, then take the remainder of its division by 15 to replace it. It is not difficult to verify that these 15 sets meet all the conditions of the problem.

In summary, the minimum value of $n$ is 15.