Very good problem.

Denote $b=\frac{AB}{AM}, c=\frac{AC}{AN}, d=\frac{AD}{AP}$. First note that $b+c+d=4$. This follows from the equations $\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}=4\overrightarrow{AG}$ and $\beta\overrightarrow{AM}+\gamma\overrightarrow{AN}+\delta\overrightarrow{AP}=\overrightarrow{AG}$ for some nonnegative $\beta, \gamma, \delta$ with $\beta+\gamma+\delta=1$. Hence $$\sum_{cyc}[BMNP]=\sum_{cyc}([ABNP]-[AMNP])=\sum_{cyc}\frac{1}{c}\frac{1}{d}\left(1-\frac{1}{b}\right)[ABCD]$$$$=[ABCD]\sum_{cyc}\frac{b-1}{bcd}=[ABCD]\frac{1}{bcd}\le [ABCD]\frac{1}{(\frac{b+c+d}{3})^3}=\frac{27}{64}[ABCD].$$This value is achieved when $MNP||BCD$. Also compare this problem with its 2D variant.