The answer is $1$, $5$ and $8$.

Checking by hand for $n = 1, 2, \ldots, 6$, we see that $1$ and $5$ work. For $n \ge 7$, $2n + 7$ should be a prime number. Because, otherwise there exists a prime divisor of $2n + 7$ which is less than or equal to $n$ since $2n + 7$ is odd, but it divides $n!$.

Now let $2n + 7 = p \ge 21$ where $p$ is a prime number. Then the condition is equivalent to $\left(\frac{p-7}{2}\right)! \equiv 1 \pmod{p}$. Wilson's theorem gives that $(p-1)! \equiv -1 \pmod{p}$. On the other hand
$$(p-1)! \equiv (-1)^{\frac{p-7}{2}} \cdot \left(\frac{p-7}{2}\right)!^2 \cdot \frac{p-5}{2} \cdot \frac{p-3}{2} \cdot \frac{p-1}{2} \cdot \frac{p+1}{2} \cdot \frac{p+3}{2} \cdot \frac{p+5}{2} \equiv (-1)^{\frac{p-1}{2}} \frac{225}{64} \pmod{p}$$
$$\text{Thus we obtain } 225 \equiv (-1)^{\frac{p+1}{2}} 64 \pmod{p}.$$
If $p \equiv 1 \pmod{4}$, then $p \mid 225 + 64 = 17^2$ but $p \ge 21$, no solution exists.
If $p \equiv 3 \pmod{4}$, then $p \mid 225 - 64 = 7 \cdot 23$. As $p \ge 21$ we have $p = 23$, that is $n = 8$ and it satisfies the condition.