Solution:

Give the cube coordinates so that $A$ is the origin, the edges lie along the axes, and the edge length is $s:$ then $B = (s, 0, 0),$ $C = (0, s, s),$ $D = (s, s, s)$ satisfy the description ($\overline{AB} \parallel \overline{CD}$ are edges and $\overline{AC},$ $\overline{BD}$ are face diagonals). Height above $\mathcal{P}$ is a linear function $h(x, y, z) = u_1 x + u_2 y + u_3 z$ for some unit vector $u.$ The plane of rectangle $ABDC$ has normal direction $(0, 1, -1),$ and perpendicularity to $\mathcal{P}$ means the vertical direction $u$ lies in that plane, so $u_2 = u_3.$ The heights of $B$ and $C$ give $s u_1 = 2$ and $s(u_2 + u_3) = 8,$ so $su_2 = su_3 = 4,$ and $|u| = 1$ forces $s^2 = 2^2 + 4^2 + 4^2 = 36.$ Thus $s = 6$ and $u = \frac{1}{3}(1, 2, 2)$ (and indeed $h(D) = 10$).

The water is the region of $[0, 6]^3$ where $h \le 7,$ i.e. $x + 2y + 2z \le 21.$ For fixed $x = a,$ the slice is $\{(y, z) \in [0,6]^2 : y + z \le \tfrac{21 - a}{2}\},$ and since $\frac{21 - a}{2}$ lies between $6$ and $12,$ its area is $36 - \frac{1}{2}\left(12 - \frac{21 - a}{2}\right)^2 = 36 - \frac{(3 + a)^2}{8}.$

Integrating, $$V = \int_0^6 \left(36 - \frac{(3 + a)^2}{8}\right) da = 216 - \frac{9^3 - 3^3}{24} = 216 - \frac{702}{24} = \frac{747}{4},$$ so $m + n = 747 + 4 = 751.$