11. Determine all polynomials of the form $x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}\left(a_{i}= \pm 1,1 \leqslant i \leqslant n\right)$, such that all zeros of each polynomial are real.
11. Let the $n$ real roots of $x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}=0$ be $x_{1}, x_{2}, \cdots, x_{n}$. On one hand, by Vieta's formulas, $x_{1}^{2}+ x_{2}^{2}+\cdots+x_{n}^{2}=\left(\sum_{i=1}^{n} x_{i}\right)^{2}-2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}=a_{1}^{2}-2 a_{2}$. On the other hand, by the arithmetic-geometric mean inequality and Vieta's formulas, we have $x_{1}^{2}+\cdots+x_{n}^{2} \geqslant n \sqrt[n]{x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2}}=n\left(a_{n}^{2}\right)^{\frac{1}{n}}$, with equality if and only if $x_{1}^{2}=x_{2}^{2}=\cdots=x_{n}^{2}$. Therefore, $a_{1}^{2}-2 a_{2} \geqslant n\left(a_{n}^{2}\right)^{\frac{1}{n}}$. Substituting $a_{i}= \pm 1$ yields $n \leqslant 1 \pm 2$, i.e., $n \leqslant 3$. Based on the above argument, the polynomials that meet the criteria are $x-1, x+1, x^{2}+x-1, x^{2}-x-1, x^{3}+x^{2}-x-1, x^{3}-x^{2}-x+1$.