What is the volume of the $3$-dimensional elliptope?

The integral can be separated:

$$I = 2\int_{-1}^1 \sqrt{1-x^2} dx \cdot \int_{-1}^1 \sqrt{1-y^2} dy = 2\left(\int_{-1}^1 \sqrt{1-t^2} dt\right)^2$$

This integral is straight-forward using the substitution $t=\sin\theta$:

$$\int_{-1}^1 \sqrt{1-t^2} dt = \int_{-\pi/2}^{\pi/2} \sqrt{1-\sin^2\theta} \cos\theta d\theta = \int_{-\pi/2}^{\pi/2} |\cos\theta|\cos\theta d\theta $$

$$=\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta = \dfrac{1}{2}\int_{-\pi/2}^{\pi/2} (1+\cos2\theta) d\theta = \dfrac{1}{2}\left(\theta + \dfrac{1}{2}\sin 2\theta\right)\Big|_{-\pi/2}^{\pi/2} = \dfrac{\pi}{2}$$

Therefore

$$I = 2\left(\int_{-1}^1 \sqrt{1-t^2} dt\right)^2 = 2\left(\dfrac{\pi}{2}\right)^2 = \dfrac{\pi^2}{2}$$

Then integrand factors as $\sqrt{(1-x^2)(1-y^2)}=\sqrt{(1-x^2)}\sqrt{(1-y^2)}$ and every factor can be integrated separately. But you recognize the integral for the area of a half circle of radius $1$, hence

$$I=2\left(\frac\pi2\right)^2.$$