Time Dilation inside a hollow shell

For an asymptotically flat metric, the proper time measured by a "stationary" observer (defined here as one whose path through spacetime only has changing $t$, and no changing spatial coordinates) is $$ d \tau = \sqrt{ - g_{tt}} dt, $$ where $g_{tt}$ is the time-time component of the metric. For a "weak" gravitational field, this works out to be $$ g_{tt} \approx - \left( 1 + \frac{2 \Phi}{c^2} \right), $$ where $\Phi$ is the gravitational potential, defined such that $\Phi \to 0$ as $r \to \infty$. Thus, $$ d \tau = \sqrt{ 1 + \frac{2 \Phi}{c^2}} dt. $$ In this form, it is pretty obvious that the time dilation factor is the same everywhere inside the shell, since $\Phi$ is a constant inside a hollow shell (compare the electrostatic equivalent if you're not convinced of this.)

Note that your formula, in terms of the escape velocity, is equivalent to this one if you define the escape velocity at any point as "the velocity for which the object's total energy is zero." (Zero total energy means, of course, that the particle can escape to infinity.) In this case, we have $$ \frac{1}{2} m v_\text{esc}^2 + m \Phi = 0 \quad \Rightarrow \quad v_\text{esc}^2 = - 2 \Phi $$ and your result above is recovered. In this interpretation, the "escape velocity" from inside a hollow sphere would be the same as the escape velocity from the surface: if we launch a projectile inside the shell, it will travel with constant velocity until it reaches the surface of the shell; and if we open a little porthole in the shell at that point for the projectile, it's as if we launched it from the surface with that same velocity.