The minimum number of participants is $17$.
Let $a_n$ be the number of participants obtaining score $n$, and let $m = a_0 + a_1 + \cdots + a_{10}$ be the total number of participants. Using the mean, we have $\sum_{n=0}^{10} n a_n = 3m$. This means
$$\sum_{n=4}^{10} (n-3)a_n = 3a_0 + 2a_1 + a_2. \qquad (1)$$

Case 1. $m$ is odd
Since $3$ is the median, we must have $a_0 + a_1 + a_2 \le \frac{m-1}{2}$. This implies
$$a_0 + a_1 + a_2 + 1 \le \sum_{n=3}^{10} a_n. \qquad (2)$$
Using (1) and (2), we obtain
$$3a_0 + 2a_1 + a_2 = \sum_{n=4}^{10} (n-3)a_n = \sum_{n=3}^{10} a_n + \sum_{n=4}^{10} (n-4)a_n - a_3 \\ \ge (a_0 + a_1 + a_2 + 1) + (3a_7) - a_3.$$
This implies $2a_0 + a_1 + a_3 \ge 3a_7 + 1$. Since the mode is $7$, we have $a_0, a_1, a_3 \le a_7 - 1$. This yields
$$3a_7 + 1 \le 2a_0 + a_1 + a_3 \le 4(a_7 - 1),$$
and hence $a_7 \ge 5$. Now, we have
$$m \ge a_0 + a_1 + a_3 + a_7 = (2a_0 + a_1 + a_3) - a_0 + a_7 \ge (3a_7 + 1) - (a_7 - 1) + a_7 \ge 17.$$
Equality holds when $a_0 = a_1 = a_3 = 4$, $a_7 = 5$ and all other $a_j$'s are $0$. One checks that all conditions are satisfied.

Case 2. $m$ is even
If $a_3 > 0$, then the two scores in the middle are $3$. The bound $a_0 + a_1 + a_2 \le \frac{m-1}{2}$ used in case 1 still holds, and so we still have $m \ge 17$.
If $a_3 = 0$, then we only have $a_0 + a_1 + a_2 \le \frac{m}{2}$. This implies
$$a_0 + a_1 + a_2 \le \sum_{n=4}^{10} a_n. \qquad (3)$$
Using (1) and (3), we obtain
$$3a_0 + 2a_1 + a_2 = \sum_{n=4}^{10} (n-3)a_n = \sum_{n=4}^{10} a_n + \sum_{n=4}^{10} (n-4)a_n \\ \ge (a_0 + a_1 + a_2) + (3a_7).$$
This implies $2a_0 + a_1 \ge 3a_7$. But then one of $a_0, a_1$ is greater than or equal to $a_7$, contradicting the fact that $7$ is the mode.
In any case, we have $m \ge 17$, and so $17$ is the minimum value.