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11. $(-3,3)$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ be two points on the given ellipse, then the equation of the perpendicular bisector $C P$ of $A B$ is $y- \frac{y_{1}+y_{2}}{2}=-\frac{x_{1}-x_{2}}{y_{1}-y_{2}}\left(x-\frac{x_{1}+x_{2}}{2}\right)$, the x-coordinate of its intersection with the x-axis is $x=\frac{y_{1}^{2}-y_{2}^{2}}{2\left(x_{1}-x_{2}\right)}+\frac{x_{1}+x_{2}}{2}=$ $\frac{y_{1}^{2}-y_{2}^{2}+x_{1}^{2}-x_{2}^{2}}{2\left(x_{1}-x_{2}\right)}$. Since $\frac{x_{1}^{2}}{16}+\frac{y_{1}^{2}}{4}=1$, $\frac{x_{2}^{2}}{16}+\frac{y_{2}^{2}}{4}=1$, hence $x_{1}^{2}-x_{2}^{2}=-4\left(y_{1}^{2}-y_{2}^{2}\right)$, so $x=\frac{3}{8}\left(x_{1}\right.$ $\left.+x_{2}\right)$, obviously; $|x_{1}+x_{2}|<8$, therefore $|x|<3$.