The smallest $n$ is $8$.
We use terminologies in set theory. Let $A_1, A_2, \dots, A_n$ be sets corresponding to the $n$ subjects, while the elements correspond to the students getting the best score in that subject. It is given that $|A_j| = 3$ and $|A_i \cap A_j| = 1$ for any $1 \le i < j \le n$. Suppose $n \ge 8$. We shall show that $|A_1 \cap A_2 \cap \dots \cap A_n| \ge 1$.
WLOG assume $1$ is an element that belongs to the most number of sets. Suppose $1 \in A_j$ for $j = 1, 2, \dots, m$ and $1 \notin A_j$ for $j = m+1, m+2, \dots, n$. Suppose on the contrary that $m < n$.
Consider $A_n = \{a, b, c\}$. For $j = 1, 2, \dots, m$, since $|A_n \cap A_j| = 1$, each $A_j$ consists of one of $a, b, c$. Also, each of $a, b, c$ belongs to at most one of these $A_j$'s, since the only common element in these sets is $1$, but $1 \notin A_n$. Thus, $m \le 3$. This means each element belongs to at most $3$ sets.
Now, suppose $A_1 = \{1, 2, 3\}$. Since each of $1, 2, 3$ belongs to at most two other sets, we must have $n \le 1 + 2 \times 3 = 7$, which is a contradiction. Therefore, $|A_1 \cap A_2 \cap \dots \cap A_n| \ge 1$.
It suffices to provide an example for $n = 7$ such that $A_1 \cap A_2 \cap \dots \cap A_n$ is empty, since we can remove some sets if $n$ is less than $7$. Indeed, consider
$$A_1 = \{1, 2, 3\},\ A_2 = \{1, 4, 5\},\ A_3 = \{1, 6, 7\},\ A_4 = \{2, 4, 6\},\ A_5 = \{2, 5, 7\},\ A_6 = \{3, 4, 7\},\ A_7 = \{3, 5, 6\}.$$
It is routine to check that $|A_i \cap A_j| = 1$ for all $i, j$, but no element belongs to all of them.