Solve $n_{\min }=41$.
First, prove that $n=41$ meets the conditions. Use proof by contradiction: Assume there exist 15 subsets of $X$ such that the union of any 7 of them contains at least 41 elements, and the intersection of any 3 of them is empty. Then each element belongs to at most 2 subsets. Without loss of generality, assume each element belongs to exactly 2 subsets (otherwise, add some elements to some subsets, and the conditions still hold). By the pigeonhole principle, there must be a subset, say $A$, containing at least $\left[\frac{56 \times 2}{15}\right]+1=8$ elements. Let the other 14 subsets be $A_{1}$, $A_{2}, \cdots, A_{14}$.

Consider any 7 subsets that do not include $A$, each corresponding to 41 elements in $X$. All 7-subset groups that do not include $A$ together correspond to at least $41 C_{14}^{7}$ elements.

On the other hand, for an element $a$, if $a \notin A$, then 2 of $A_{1}, A_{2}, \cdots, A_{14}$ contain $a$, so $a$ is counted $\left(C_{14}^{7}-C_{12}^{7}\right)$ times; if $a \in A$, then 1 of $A_{1}, A_{2}, \cdots, A_{14}$ contains $a$, so $a$ is counted $\left(C_{14}^{7}-C_{13}^{7}\right)$ times. Thus, $41 C_{14}^{7} \leqslant (56-|A|)\left(C_{14}^{7}-C_{12}^{7}\right) + |A| \cdot \left(C_{14}^{7}-C_{13}^{7}\right) = 56\left(C_{14}^{7}-C_{12}^{7}\right) - |A|\left(C_{13}^{7}-C_{12}^{7}\right) \leqslant 56\left(C_{14}^{7}-C_{12}^{7}\right) - 8\left(C_{13}^{7}-C_{12}^{7}\right)$, which simplifies to $48 C_{12}^{7} + 8 C_{13}^{7} \leqslant 15 C_{14}^{7}$. Simplifying further, we get $3 \times 48 + 4 \times 13 \leqslant 15 \times 13$, or $196 \leqslant 195$, which is a contradiction.

Next, prove that $n \geqslant 41$ using proof by contradiction.
Assume $n \leqslant 40$, and let $X=\{1,2, \cdots, 56\}$. Define $A_{i}=\{x \in X \mid x \equiv i(\bmod 7)\}$ $(i=1,2, \cdots, 7)$ and $B_{j}=\{x \in X \mid x \equiv j(\bmod 8)\}$ $(j=1,2, \cdots, 8)$. Clearly, $|A_{i}|=8$, $|A_{i} \cap A_{j}|=0$ for $1 \leqslant i<j \leqslant 7$, $|B_{j}|=7$, and $|B_{i} \cap B_{j}|=0$ for $1 \leqslant i<j \leqslant 8$. Additionally, by the Chinese Remainder Theorem, $|A_{i} \cap B_{j}|=1$ for $1 \leqslant i \leqslant 7$ and $1 \leqslant j \leqslant 8$. Thus, for any 3 subsets, there must be 2 that are both $A_{i}$ or both $B_{j}$, and their intersection is empty. For any 7 subsets, let there be $t$ (where $0 \leqslant t \leqslant 7$) of them be $A_{i}$ and $7-t$ be $B_{j}$. By the principle of inclusion-exclusion, the number of elements in the union of these 7 subsets is $8t + 7(7-t) - t(7-t) = 49 - t(6-t) \geqslant 49 - 9$ (since $0 \leqslant t \leqslant 7$) $= 40$. Therefore, the union of any 7 subsets contains at least 40 elements, but the intersection of any 3 subsets is empty, so $n \geqslant 41$.

In conclusion, the minimum value of $n$ is 41.