As shown in the figure, sphere $O$ is the inscribed sphere of the regular tetrahedron $P-DEF$, the midpoint of $AC$ is $G$, and $OM \perp PG$ at $M$. Then $r=1 \Rightarrow PO=3 \Rightarrow PM=2\sqrt{2}=\frac{1}{3}PG$, similarly $AN=2\sqrt{2}=\frac{1}{3}AF \Rightarrow MN=2\sqrt{6}$.
Thus, the region within one face of the tetrahedron that the small sphere can touch is a regular triangle with side length $MN$ and its interior, whose area is $\frac{1}{4}$ of the area of one face of the tetrahedron.

Therefore, the area of the container's inner wall that the small sphere can never touch is $\frac{3}{4}$ of the surface area of the tetrahedron, i.e., $S=\frac{3}{4} \times 4 \times \frac{\sqrt{3}}{4} \times (4\sqrt{6})^2=72\sqrt{3}$.
12. A small ball with a radius of 1 can move freely in all directions inside a regular tetrahedron container with an internal edge length of $4 \sqrt{6}$. The area of the container's inner wall that the ball can never touch is $\qquad$.