As shown in Fig. 1, consider the situation where the ball is in a corner of the container. Draw the plane $A_1B_1C_1 \parallel ABC$, tangent to the ball at point $D$. Then the ball center $O$ is also the center of the tetrahedron $P-A_1B_1C_1$, with $PO \perp A_1B_1C_1$ and the foot point $D$ being the center of $\triangle A_1B_1C_1$.

Since

$$\begin{aligned} V_{P-A_1B_1C_1} &= \frac{1}{3} \times S_{\triangle A_1B_1C_1} \times PD \\ &= 4 \times V_{O-A_1B_1C_1} \\ &= 4 \times \frac{1}{3} \times S_{\triangle A_1B_1C_1} \times OD, \end{aligned}$$
we have $PD = 4OD = 4r$, where $r$ is the radius of the ball. It follows that
$$PO = PD - OD = 3r.$$
Suppose that the ball is tangent to the plane $PAB$ at point $P_1$. Then we have
$$PP_1 = \sqrt{PO^2 - OP_1^2} = \sqrt{(3r)^2 - r^2} = 2\sqrt{2}r.$$
As shown in Fig. 2, it is easy to see that the locus of the ball on the plane $PAB$ is also a regular triangle, denoted by $P_1EF$. Through $P_1$ draw $P_1M \perp PA$ with point $M$ on $PA$. Then $\angle MPP_1 = \frac{\pi}{6}$, and
$$PM = PP_1 \times \cos \angle MPP_1 = 2\sqrt{2}r \times \frac{\sqrt{3}}{2} = \sqrt{6}r.$$

It follows that $P_1E = PA - 2PM = a - 2\sqrt{6}r$, where $a = PA$. Now, the space on $PAB$ which the ball will never touch is the shaded part of Fig. 2, and its size is equal to
$$\begin{align*} S_{\triangle PAB} - S_{\triangle P_1EF} &= \frac{\sqrt{3}}{4}(a^2 - (a - 2\sqrt{6}r)^2) \\ &= 3\sqrt{2}ar - 6\sqrt{3}r^2 \\ &= 24\sqrt{3} - 6\sqrt{3} \\ &= 18\sqrt{3}, \end{align*}$$
since $r = 1$ and $a = 4\sqrt{6}$ under given conditions. Then the total untouched area is
$$4 \times 18\sqrt{3} = 72\sqrt{3}.$$