Solution:

Each radicand factors: $2x - xy = x(2 - y),$ and so on, so $0 \lt x, y, z \le 2.$ Substitute $x = 2\sin^2\alpha,$ $y = 2\sin^2\beta,$ $z = 2\sin^2\gamma$ with $\alpha, \beta, \gamma \in \left(0^\circ, 90^\circ\right].$ Then $\sqrt{x(2 - y)} = \sqrt{4\sin^2\alpha\cos^2\beta} = 2\sin\alpha\cos\beta,$ and each equation collapses by the sine addition formula: $$2\sin(\alpha + \beta) = 1, \qquad 2\sin(\beta + \gamma) = \sqrt{2}, \qquad 2\sin(\gamma + \alpha) = \sqrt{3}.$$

Taking $\alpha + \beta = 30^\circ,$ $\beta + \gamma = 45^\circ,$ $\gamma + \alpha = 60^\circ$ and solving, $\alpha = 22.5^\circ,$ $\beta = 7.5^\circ,$ $\gamma = 37.5^\circ.$ (The supplementary branch choices consistent with the angle ranges lead to the same value of the final square.) By the double-angle identity, $1 - x = \cos 2\alpha = \cos 45^\circ,$ $1 - y = \cos 15^\circ,$ and $1 - z = \cos 75^\circ.$

Therefore $$(1 - x)(1 - y)(1 - z) = \frac{\sqrt{2}}{2}\cos 15^\circ \sin 15^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sin 30^\circ}{2} = \frac{\sqrt{2}}{8},$$ whose square is $\frac{2}{64} = \frac{1}{32}.$ Thus $m + n = 1 + 32 = 33.$