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We'll denote with ${\cal I}$ the requested answer.
\begin{align}
{\cal I}&\equiv
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\Theta\pars{4 - x^{2} - y^{2} - z^{2}}\Theta\pars{1 - 2x^{2} - z^{2}}\,
\dd x\,\dd y\,\dd z
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Theta\pars{1 - 2x^{2} - z^{2}}\,\dd x\,\dd z\int_{-\infty}^{\infty}\Theta\pars{\bracks{4 - x^{2} - z^{2}} - y^{2}}\,\dd y
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Theta\pars{1 - 2x^{2} - z^{2}}
\Theta\pars{4 - x^{2} - z^{2}}\,\dd x\,\dd z
\int_{-\root{\vphantom{\Large A}4- x^{2} - z^{2}}}
^{\root{\vphantom{\Large A}4- x^{2} - z^{2}}}\,\dd y
\\[3mm]&=
2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Theta\pars{1 - 2x^{2} - z^{2}}
\Theta\pars{4 - x^{2} - z^{2}}\root{\vphantom{\Large A}4- x^{2} - z^{2}}
\,\dd x\,\dd z
\\[3mm]&=
16\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Theta\pars{1 - 8x^{2} - 4z^{2}}
\Theta\pars{1 - x^{2} - z^{2}}\root{\vphantom{\Large A}1- x^{2} - z^{2}}
\,\dd x\,\dd z
\end{align}
Whith the change of variables $x \equiv \rho\cos\pars{\theta}$,
$z \equiv \rho\sin\pars{\theta}$ $\pars{~0 \leq \rho\,,\ 0 \leq \theta < 2\pi~}$, we'll get
\begin{align}
{\cal I}&=
16\int_{0}^{1}\dd\rho\,\rho\root{1- \rho^{2}}
\int_{0}^{2\pi}
\Theta\pars{1 - 4\rho^{2}\bracks{2\cos^{2}\pars{\theta} - \sin^{2}\pars{\theta}}}
\,\dd\theta
\\[3mm]&=
8\int_{0}^{1}\dd\rho\,\root{1 - \rho}\int_{0}^{2\pi}
\Theta\pars{1 - 8\rho + 12\rho\sin^{2}\pars{\theta}}\,\dd\theta
\\[3mm]&=
8\int_{0}^{1}\dd\rho\,\root{1 - \rho}
\int_{0}^{2\pi}
\Theta\pars{\sin^{2}\pars{\theta} - {8\rho - 1 \over 12\rho}}\,\dd\theta
\\[3mm]&=
16\pi\int_{0}^{1/8}\root{1 - \rho}\,\dd\rho
+
8\int_{1/8}^{1}\dd\rho\,\root{1 - \rho}
\int_{0}^{2\pi}
\Theta\pars{\sin^{2}\pars{\theta} - {8\rho - 1 \over 12\rho}}\,\dd\theta\tag{1}
\end{align}
Let's perform the angular $\pars{~\mbox{over}\ \theta~}$ integration
$\pars{~\mbox{notice that}\ \bracks{8\rho - 1}/\bracks{12\rho} < 1\ \mbox{when}\
\rho > 0 ~}$:
\begin{align}
&\int_{0}^{2\pi}
\Theta\pars{\sin^{2}\pars{\theta} - {8\rho - 1 \over 12\rho}}\,\dd\theta
=
4\int_{0}^{\pi/2}
\Theta\pars{\sin^{2}\pars{\theta} - {8\rho - 1 \over 12\rho}}\,\dd\theta
\\[3mm]&=
4\int_{\arcsin\pars{\root{\vphantom{\huge A}\bracks{8\rho - 1}/\bracks{12\rho}}}}^{\pi/2}\dd\theta
=
4\bracks{{\pi \over 2} - \arcsin\pars{\root{8\rho - 1 \over 12\rho}}}
\\[3mm]&=
2\pi - 4\arcsin\pars{\root{8\rho - 1 \over 12\rho}}
\end{align}
By replacing this result in $\pars{1}$, we'll get:
\begin{align}
{\cal I}&=
16\pi\int_{0}^{1}\root{1 - \rho}\,\dd\rho
-
32\int_{1/8}^{1}\dd\rho\,\root{1 - \rho}\arcsin\pars{\root{8\rho - 1 \over 12\rho}}
\\[3mm]&=
24\pi - 32\underbrace{\int_{1/8}^{1}\root{1 - \rho}
\arcsin\pars{\root{{2 \over 3} - { 1 \over 12\rho}}}\,\dd\rho}
_{\ds{\approx\ 0.387853}}
\end{align}
$$
\color{#0000ff}{\large{\cal I} = 62.9869}
$$
