Answer: 544

Solution:

Scale to a semicircle of radius $1$ and ask for the smallest square containing it when its diameter makes angle $\theta$ with one pair of sides, where $0 \le \theta \le 90^\circ.$ Squeeze the semicircle between two pairs of parallel lines in the square's two side directions: in each direction one line of the pair is tangent to the arc and the other passes through an endpoint of the diameter, and the distances between the pairs are $1 + \cos\theta$ and $1 + \sin\theta.$ So the smallest enclosing square in that orientation has side $\max\{1 + \cos\theta,\ 1 + \sin\theta\},$ which is minimized when $\theta = 45^\circ,$ giving side $1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}.$

Scaling this optimal configuration so the square has side $8,$ the radius becomes $r = \frac{8}{(2 + \sqrt{2})/2}$ $= \frac{16}{2 + \sqrt{2}}$ $= 8\left(2 - \sqrt{2}\right),$ so $$\begin{aligned} d &= 2r = 16\left(2 - \sqrt{2}\right) \\ &= 32 - 16\sqrt{2} = 32 - \sqrt{512}. \end{aligned}$$

Thus $m + n = 32 + 512 = 544.$