25. First, find the maximum value. By the AM-GM inequality, we have $\sqrt[n]{x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}} \leqslant \frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}=\frac{1}{n}$. The equality holds when $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}>\frac{1}{n}(n \geqslant 2)$. Therefore, the maximum value is $n^{-\frac{n}{2}}$. Next, find the minimum value. Let $y_{1}=x_{1}, \cdots, y_{n-2}=x_{n-2}, y_{n-1}=\sqrt{x_{n-1}^{2}+x_{n}^{2}-\frac{1}{n^{2}}}, y_{n}=\frac{1}{n}$, then $y_{i} \geqslant \frac{1}{n}, i=1,2, \cdots, n$, and $y_{1}^{2}+\cdots+y_{n}^{2}=x_{1}^{2}+\cdots+x_{n}^{2}=1$. Since $y_{n-1}^{2} y_{n}^{2}-x_{n-1}^{2} x_{n}^{2}=-\left(x_{n-1}^{2}-\frac{1}{n^{2}}\right)\left(x_{n}^{2}-\frac{1}{n^{2}}\right) \leqslant 0$, we have $y_{1} y_{2} \cdots y_{n-2} y_{n-1} y_{n} \leqslant x_{1} x_{2} \cdots x_{n-2} x_{n-1} x_{n}$. Repeating this process $n-1$ times, we get $x_{1} x_{2} \cdots x_{n} \geqslant\left(\frac{1}{n}\right)^{n-1} \sqrt{1-\frac{n-1}{n^{2}}}=\frac{\sqrt{n^{2}-n+1}}{n^{n}}$. The equality holds when $x_{1}=\cdots=x_{n-1}=\frac{1}{n}$, $x_{n}=\frac{\sqrt{n^{2}-n+1}}{n}$. Therefore, the minimum value is $\frac{\sqrt{n^{2}-n+1}}{n^{n}}$.