Convert the point $(4, 4, 4 \sqrt{6})$ in rectangular coordinates to spherical coordinates.  Enter your answer in the form $(\rho,\theta,\phi),$ where $\rho > 0,$ $0 \le \theta < 2 \pi,$ and $0 \le \phi \le \pi.$
We have that $\rho = \sqrt{4^2 + 4^2 + (4 \sqrt{6})^2} = 8 \sqrt{2}.$  We want $\phi$ to satisfy
\[4 \sqrt{6} = 8 \sqrt{2} \cos \phi,\]so $\phi = \frac{\pi}{6}.$

We want $\theta$ to satisfy
\begin{align*}
4 &= 8 \sqrt{2} \sin \frac{\pi}{6} \cos \theta, \\
4 &= 8 \sqrt{2} \sin \frac{\pi}{6} \sin \theta.
\end{align*}Thus, $\theta = \frac{\pi}{4},$ so the spherical coordinates are $\boxed{\left( 8 \sqrt{2}, \frac{\pi}{4}, \frac{\pi}{6} \right)}.$