Put the corner at the origin $O$ with the two sides along the positive axes, and write $a = \sqrt{17}.$ The cut on the bottom edge starts at $P = (a, 0),$ and the two cuts meet at $R$ on the diagonal $y = x,$ each making a $30^\circ$ angle with the diagonal. In triangle $OPR,$ $\angle ROP = 45^\circ$ and $\angle ORP = 30^\circ,$ so the Law of Sines gives $PR = \frac{OP\sin 45^\circ}{\sin 30^\circ} = a\sqrt{2}.$ The fold lines are the horizontal and vertical lines through $R.$ Let $S$ be the point of the horizontal fold line directly above $P,$ and $T = (a, a)$ the point where the vertical line through $P$ meets the diagonal. Since $\angle OPR = 105^\circ,$ segment $PR$ makes a $75^\circ$ angle with the bottom edge, so $$\begin{aligned} SP &= PR\sin 75^\circ \\ &= a\sqrt{2} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} \\ &= a\,\frac{\sqrt{3} + 1}{2}, \\ ST &= SP - PT \\ &= a\,\frac{\sqrt{3} + 1}{2} - a \\ &= a\,\frac{\sqrt{3} - 1}{2}. \end{aligned}$$

When the bottom strip folds up along the horizontal line through $R,$ point $P$ stays at distance $SP$ from $S,$ moving in the vertical plane through $P$ perpendicular to that fold line. By symmetry the two taped cut edges meet above the diagonal, so $P$ lands at a point $P'$ directly above $T,$ and $P'T$ is the height of the tray. By the Pythagorean theorem, $$\begin{aligned} P'T^2 &= P'S^2 - ST^2 \\ &= a^2\left(\frac{\sqrt{3} + 1}{2}\right)^2 \\ &\quad {}- a^2\left(\frac{\sqrt{3} - 1}{2}\right)^2 \\ &= a^2\sqrt{3}. \end{aligned}$$

So the height is $a \cdot 3^{1/4}$ $= \sqrt{17} \cdot \sqrt[4]{3}$ $= \sqrt[4]{17^2 \cdot 3}$ $= \sqrt[4]{867},$ and $m + n = 867 + 4 = 871.$