Answer: 80

Solution:

Place the base at $A = (0,0,0),$ $B = (4,0,0),$ $C = (4,4,0),$ $D = (0,4,0);$ the apex is then $E = (2, 2, 2\sqrt{2}),$ since $2^2 + 2^2 + 8 = 16.$ The given midpoints are $R = (1, 1, \sqrt{2}),$ $S = (4, 2, 0),$ and $T = (2, 4, 0),$ and all three satisfy $x + y + 2\sqrt{2}\,z = 6,$ the equation of the cutting plane.

Parametrizing edges $\overline{BE}$ and $\overline{DE}$ shows the plane meets them at $U = \left(\frac{7}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}\right)$ and $V = \left(\frac{1}{2}, \frac{7}{2}, \frac{\sqrt{2}}{2}\right).$ The cross-section is the pentagon $RUSTV$ with $RU = RV = \sqrt{7},$ $US = VT = \sqrt{3},$ $ST = 2\sqrt{2},$ and diagonal $UV = 3\sqrt{2}.$

Split the pentagon along $\overline{UV}.$ Isosceles triangle $RUV$ has height $\sqrt{7 - \frac{9}{2}} = \sqrt{\frac{5}{2}}$ and area $\frac{1}{2} \cdot 3\sqrt{2} \cdot \sqrt{\frac{5}{2}} = \frac{3\sqrt{5}}{2}.$ Isosceles trapezoid $USTV$ has height $\sqrt{3 - \frac{1}{2}} = \sqrt{\frac{5}{2}}$ and area $\frac{1}{2}(3\sqrt{2} + 2\sqrt{2})\sqrt{\frac{5}{2}} = \frac{5\sqrt{5}}{2}.$ The total is $4\sqrt{5} = \sqrt{80},$ so $p = 80.$