Suppose there exist 15 subsets of $X$, the union of every 7 of these 15 subsets contains at least 41 elements, and there exists no 3 of the 15 subsets whose intersection is nonempty.
Since every element of $X$ can only belong 2 of the 15 subsets, we can suppose that every element of $X$ belongs to exactly 2 of the 15 subsets. Otherwise we can add a few more elements to some sets of the 15 subsets, and the condition still holds.
By the Pigeonhole Principle, there is a set of the 15 subsets (suppose it is $A$), such that $|A| \ge \lfloor \frac{2 \times 56}{15} \rfloor + 1 = 8$. Let the other 14 sets be $A_1, A_2, \dots, A_{14}$.
The union of every 7 sets of $A_1, A_2, \dots, A_{14}$ contains at least 41 elements. So the total number $y \le 41C_{14}^7$.

We use another way to compute the value of $y$.
For $a \in X$, if $a \notin A$, then $a$ belongs to exactly 2 sets of $A_1, A_2, \dots, A_{14}$. So $a$ is counted $C_{14}^7 - C_{12}^7$ times. If $a \in A$, then $a$ belongs to only one set of $A_1, A_2, \dots, A_{14}$. So $a$ is counted $C_{14}^7 - C_{13}^7$ times. It follows that
$$\begin{aligned} 41C_{14}^7 \le y \le (56 - |A|)(C_{14}^7 - C_{12}^7) + |A|(C_{14}^7 - C_{13}^7) \\ = 56(C_{14}^7 - C_{12}^7) - |A|(C_{13}^7 - C_{12}^7) \\ \le 56(C_{14}^7 - C_{12}^7) - 8(C_{13}^7 - C_{12}^7), \end{aligned}$$
that is, $196 \le 195$, a contradiction.

Next, we prove $n \ge 41$.
If $n \le 40$, suppose $X = \{1, 2, \dots, 56\}$. let
$$A_i = \{i, i+7, i+14, i+21, i+28, i+35, i+42, i+49\},$$
$$B_j = \{j, j+8, j+16, j+24, j+32, j+40, j+48\},$$
It is easy to see that
$$\begin{aligned} |A_i| &= 8 \quad (i = 1, 2, \dots, 7), & |A_i \cap A_j| &= 0 \quad (1 \le i < j \le 7), \\ |B_j| &= 7 \quad (j = 1, 2, \dots, 8), & |B_i \cap B_j| &= 0 \quad (1 \le i < j \le 8), \\ |A_i \cap B_j| &= 1 \quad (1 \le i \le 7, 1 \le j \le 8). \end{aligned}$$
For every 3 of the 15 subsets, there are 2 sets both being $A_i$, or both being $B_j$. So the intersection of the 3 sets is empty.
But for every 7 of the 15 subsets, for example,
$$A_{i_1}, A_{i_2}, \dots, A_{i_s}, B_{j_1}, B_{j_2}, \dots, B_{j_t} \quad (s+t=7),$$
we have
$$\begin{aligned} & |A_{i_1} \cup A_{i_2} \cup \dots \cup A_{i_s} \cup B_{j_1} \cup B_{j_2} \cup \dots \cup B_{j_t}| \\ & = |A_{i_1}| + |A_{i_2}| + \dots + |A_{i_s}| + |B_{j_1}| + |B_{j_2}| + \dots + |B_{j_t}| - st \end{aligned}$$
$$\begin{aligned} &= 8s + 7t - st = 8s + 7(7 - s) - s(7 - s) \\ &= (s - 3)^2 + 40 \ge 40. \end{aligned}$$
So $n \ge 41$.