You can splitSplit up the cube into $6$ tetrahedral chunks, so the integral is the total of the integrals over each tetrahedron, which are easier to set uptetrahedral chunks. One such chunk involves the tetrahedron withhas vertices $(0,0,0),(1,0,0),(1,1,0),(1,1,1)$$\{(0,0,0),(0,0,1),(1,0,1),(1,1,1)\}$ and is enclosed by the planes $y=0$, $z=1$, $y=x$, and $z=x$:
$$\int_0^{\pi/4}\int_{\tan^{-1}(\csc\theta)}^{\pi/2}\int_0^{\sec\theta\csc\varphi}\rho^3\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta=\frac{6\sqrt3-\pi+6\ln(7+4\sqrt3)}{144}\approx0.1600976746$$$$\begin{align*} I_1 &= \int_0^\tfrac\pi4\int_0^{\arctan(\sec\theta)}\int_0^{\sec\varphi} \rho^3\sin\varphi \,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta \\ &= \frac14 \int_0^\tfrac\pi4 \int_0^{\arctan(\sec\theta)} \tan\varphi \sec^3\varphi \, d\varphi \, d\theta \\ &= \frac14 \int_0^\tfrac\pi4 \int_1^{\sqrt{1+\sec^2\theta}} f^2 \, df \, d\theta & f=\sec\varphi \\ &= \frac1{12} \int_0^\tfrac\pi4 \left(2+\tan^2\theta\right)^{3/2} \,d\theta - \frac\pi{48} \\ &= \frac13 \int_0^{\operatorname{arcsch}\sqrt2} \frac{\cosh^4u}{\cosh(2u)} \, du - \frac\pi{48} & \tan\theta = \sqrt2\,\sinh u \\ &= \frac{4\operatorname{arcsch}\sqrt2 - \pi}{48} + \frac1{12} \int_0^{\operatorname{arcsch}\sqrt2} \left(2\cosh^2u + \frac{1}{2\cosh^2u-1}\right) \, du & \text{partial fractions} \\ &= \frac{2\sqrt3 + 8\operatorname{arcsch}\sqrt2 - \pi}{48} + \frac1{12} \int_0^{\operatorname{arcsch}\sqrt2} \frac{\operatorname{sech}^2u}{1+\tanh^2u} \, du \\ &= \frac{6\sqrt3 + 24\operatorname{arcsch}\sqrt2 - \pi}{144} \end{align*}$$
TheBy symmetry, the remaining integrals' bounds in spherical coordinates are
$$\int_0^{\pi/4}\int_{\tan^{-1}(\sec\theta)}^{\tan^{-1}(\csc\theta)}\int_0^{\sec\theta\csc\varphi}$$
(tetrahedron with verticesintegrals $(0,0,0),(1,0,0),(1,0,1),(1,1,1)$)
$$\int_0^{\pi/4}\int_0^{\tan^{-1}(\sec\theta)}\int_0^{\sec\varphi}$$
(tetrahedron with vertices$I_2$ through $(0,0,0),(0,0,1),(1,0,1),(1,1,1)$)
$$\int_{\pi/4}^{\pi/2}\int_{\tan^{-1}(\sec\theta)}^{\pi/2}\int_0^{\csc\theta\csc\varphi}$$
(tetrahedron with vertices$I_5$ $(0,0,0),(0,1,0),(1,1,0),(1,1,1)$(listed below) share the same value as $I_1$, and the closed form for OP's integral follows.
$$\int_{\pi/4}^{\pi/2}\int_{\tan^{-1}(\csc\theta)}^{\tan^{-1}(\sec\theta)}\int_0^{\csc\theta\csc\varphi}$$$$\begin{array}{l|l|l} i & \text{vertices} & \text{integration range} \\ \hline 2 & \{(0,0,0),(1,0,0),(1,0,1),(1,1,1)\} & \displaystyle \int_0^\tfrac\pi4 \int_{\arctan(\sec\theta)}^{\arctan(\csc\theta)} \int_0^{\sec\theta\csc\varphi} \\ 3 & \{(0,0,0),(1,0,0),(1,1,0),(1,1,1)\} & \displaystyle \int_0^\tfrac\pi4 \int_{\arctan(\csc\theta)}^\tfrac\pi2 \int_0^{\sec\theta\csc\varphi}\\ 4 & \{(0,0,0),(0,1,0),(1,1,0),(1,1,1)\} & \displaystyle\int_\tfrac\pi4^\tfrac\pi2\int_{\arctan(\sec\theta)}^\tfrac\pi2\int_0^{\csc\theta\csc\varphi} \\ 5 & \{(0,0,0),(0,1,0),(0,1,1),(1,1,1)\} & \displaystyle\int_\tfrac\pi4^\tfrac\pi2\int_{\arctan(\csc\theta)}^{\arctan(\sec\theta)}\int_0^{\csc\theta\csc\varphi} \\ 6 & \{(0,0,0),(0,0,1),(1,1,0),(1,1,1)\} & \displaystyle\int_\tfrac\pi4^\tfrac\pi2\int_0^{\arctan(\csc\theta)}\int_0^{\sec\varphi} \end{array}$$
(tetrahedron with vertices Replacing $(0,0,0),(0,1,0),(0,1,1),(1,1,1)$)$\theta$ by $\dfrac\pi2-\theta$ reveals $I_{1,2,3}$ are respectively duplicated by $I_{6,5,4}$. Fubini's theorem $(\star)$ can in turn be used to express $I_{1,2,3}$ in terms of a common integral.
$$\int_{\pi/4}^{\pi/2}\int_0^{\tan^{-1}(\csc\theta)}\int_0^{\sec\varphi}$$$$\begin{align*} 4I_1 &= \int_0^\tfrac\pi4 \int_0^{\arctan(\sec\theta)} \sec^4\varphi \sin\varphi \, d\varphi \, d\theta \\ &= \int_0^\tfrac\pi4 \int_0^1 f\sec^2\theta\sqrt{1+f^2\sec^2\theta} \, df \, d\theta & f=\cos\theta\tan\varphi \\ &= \underbrace{\int_0^1 \int_0^1 f\sqrt{1+f^2\left(1+t^2\right)} \, df \, dt}_{\mathcal I} & t=\tan\theta \\[2ex] 4I_3 &= \int_0^\tfrac\pi4 \int_{\arctan(\csc\theta)}^\tfrac\pi2 \sec^4\theta \csc^3\varphi \, d\varphi \, d\theta \\ &= \int_0^\tfrac\pi4 \int_0^1 \sec^4\theta\sin\varphi\sqrt{1+f^2\sin^2\theta} \, df \, d\theta & f=\sin\theta\cot\varphi \\ &= \int_0^1 \int_0^1 t \sqrt{1+t^2\left(1+f^2\right)} \, df \, dt & t=\tan\theta \\ &= \int_0^1 \int_0^t \sqrt{1+f^2+t^2} \, df \, dt & f\to\frac ft \\ &= \int_0^1 \int_f^1 \sqrt{1+f^2+t^2} \, dt \, df & (\star) \\ &= \int_0^1 \int_1^\tfrac1f f\sqrt{1+f^2\left(1+t^2\right)} \, dt \, df & t\to ft\\ &= \int_0^1 \int_f^1 \frac f{t^3} \sqrt{t^2\left(1+f^2\right)+f^2} \, dt \, df & t\to\frac 1t \\ &= \int_0^1 \int_0^t \frac f{t^3} \sqrt{t^2\left(1+f^2\right)+f^2} \, df \, dt & (\star) \\ &= \mathcal I & f\to ft \\[2ex] 4I_2 &= \int_0^\tfrac\pi4 \int_{\arctan(\sec\theta)}^{\arctan(\csc\theta)} \sec^4\theta\csc^3\varphi \, d\varphi \, d\theta \\ &= \int_0^\tfrac\pi4 \int_{\sec\theta}^{\csc\theta} \sec^4\theta \frac{\sqrt{1+\varphi^2}}{\varphi^3} \, df\, d\theta & f=\tan\varphi \\ &= \int_0^\tfrac\pi4 \int_1^{\cot\theta}\frac{\sec^2\theta}{f^3}\sqrt{1+f^2\sec^2\theta} \, d\theta \, df \\ &= \int_0^\tfrac\pi4 \int_{\tan\theta}^1 \sec^2\theta \sqrt{f^2+\sec^2\theta} \, d\theta \, df \\ &= \int_0^1 \int_t^1 \sqrt{1+f^2+t^2} \, df\,dt & t=\tan\theta \\ &= \int_0^1 \int_0^f \sqrt{1+f^2+t^2} \, dt\,df & (\star) \\ &= \mathcal I & t\to ft \end{align*}$$
(tetrahedron with vertices $(0,0,0),(0,0,1),(1,1,0),(1,1,1)$)
By symmetry, the value of the original integral is$$\implies \sum_{i=1}^6 I_i = \boxed{\frac{6\sqrt3-\pi+12\log\left(2+\sqrt3\right)}{24}}$$
$$\frac{6\sqrt3-\pi+6\ln(7+4\sqrt3)}{24}\approx0.9605919565$$