WLOG let $1 \le x_1 \le x_2 \le\dots\le x_n \le 160$. Let $y_i = \sqrt{x_i}$ for all $i = 1, 2,\dots, n$. Then $1 \le y_1 \le y_2 \le\dots\le y_n \le \sqrt{160}$. Note that $x^{2}_{i} + x^{2}_{j} + x^{2}_{k} \ge 2(x_ix_j + x_jx_k + x_kx_i) \iff (\sqrt{x_i} + \sqrt{x_j} + \sqrt{x_k})(-\sqrt{x_i} + \sqrt{x_j} + \sqrt{x_k})(\sqrt{x_i} - \sqrt{x_j} + \sqrt{x_k})(\sqrt{x_i} + \sqrt{x_j} - \sqrt{x_k}) \le 0$, so $\sqrt{x_i} + \sqrt{x_j} \le \sqrt{x_k} \iff y_i + y_j \le y_k$ for all $i < j < k$. Since $y_1 = y_2 \ge 1$, we have $y_3 \ge 1 + 1 = 2, y_4 \ge 2 + 1 = 3, y_5 \ge 2 + 3 = 5, y_6 \ge 3 + 5 = 8, y_7 \ge 5 + 8 = 13 > \sqrt{160}$, so $n \le 6$. We see that $(x_1, x_2, x_3, x_4, x_5, x_6) = (1, 1, 4, 9, 25, 64)$ satisfies the conditions, so $\boxed{n = 6}$ is the maximum possible value of $n$.