$\newcommand\R{\mathbb R}$Now concerning$\newcommand\R{\mathbb R}\renewcommand{\th}{\theta}\newcommand{\om}{\omega}\newcommand{\Ga}{\Gamma}$Here is a solution for any dimension $m$$m\ge2$:
Without loss of generality (wlog), we have two open balls: $B_1=B_0(r_1)$ and $B_2=B_D(r_2)$, for some $D\in\R^m$ with $\|D\|=d$, where $B_x(r)$ is the open ball in $\R^m$ of radius $r$ centered at $x$ and $\|\cdot\|$ is the Euclidean norm.
Now it remains to integrate $\mathcal A(C,u)$ over all $u$ on the $(m-1)$-dimensional unit sphere $S^{m-1}$. When doing that, by the spherical symmetry we can replace, and will, assume that $D\cdot u$ in the above expressions for$D=d\,e_1$, where $r_{2,u}$ and$e_1$ is the first standard basis vector in $d_u$ by$\R^m$.
Then $u_1d$$D\cdot u=u_1d$, where $u_1$ is the first coordinate of the unit vector $u$, and hence $d_u=d(1-u_1^2)$. So,
$$\mathcal A(C,u)=g(u_1^2),$$\begin{equation*}
\mathcal A(C,u)=g(u_1^2), \tag{00}\label{00}
\end{equation*}
for a certain function $g=g_{r_1,r_2,d}$, expressed explicitly via an incomplete beta function. So, the étendue measure of $C$ is
$$\frac12\,\int_{S^{m-1}}\mathcal A(C,u)\,du=
\frac12\,\int_{S^{m-1}}g(u_1^2)\,du \\
=\frac12\,Eg(U_1^2)=\frac{m-1}2\,\int_0^1 g_{r_1,r_2,d}(t)(1-t)^{m-2}\,dt,$$\begin{multline}
E=E_m(r_1,r_2,d)=\frac12\,\int_{S^{m-1}}du\,\mathcal A(C,u)=
\frac12\,\int_{S^{m-1}}du\,g(u_1^2) \\
=\frac12\,|S^{m-1}|Eg(U_1^2)=b_m\,\int_0^1 dt\, g(t)t^{1/2-1}(1-t)^{(m-1)/2-1},
\tag{10}\label{10}
\end{multline}
where $|S^{m-1}|$ is the $(m-1)$-dimensional "surface area" of $S^{m-1}$ and
$U_1^2$ is a (real-valued) random variable with the beta distributionbeta distribution with parameters $1/2,(m-1)/2$, and
\begin{equation*}
b_m:=\frac{\pi^{(m-1)/2}}{\Ga((m-1)/2)}. \tag{20}\label{20}
\end{equation*}
Let us detail the latter expression in \eqref{10}. By \eqref{00}, $g(t)$ is the $(m-1)$-volume of the intersection of two balls, say $B_{1,t}$ and $B_{2,t}$, in $\R^{m-1}$, of respective radii $r_1$ and $r_2$, with parametersthe centers of the two balls at distance $1,m-1$$d_u=d\sqrt{1-t}$ from each other. Here and in what follows, $t\in(0,1)$. Let $L_k$ denote the Lebesgue measure over $\R^k$.
Without loss of generality, $r_1\ge r_2$.
There are three possible cases here:
In Case 1, $r_1+r_2\le d\sqrt{1-t}=d_u$ and hence the interiors of the balls $B_{1,t}$ and $B_{2,t}$ do not intersect, so that .
\begin{equation*}
g(t)=0 \tag{g: Case 1}\label{cs1}
\end{equation*}
In Case 2, $r_1-r_2\ge d\sqrt{1-t}=d_u$ and hence $B_{1,t}\supseteq B_{2,t}$, so that
\begin{equation*}
g(t)=L_{m-1}(B_{2,t})=\om_{m-1}r_2^{m-1} \tag{g: Case 2}\label{cs2}
\end{equation*}
where
\begin{equation}
\om_k:=\frac{\pi^{k/2}}{\Ga(k/2+1)}, \tag{30}\label{30}
\end{equation}
the $k$-volume of the unit ball in $\R^k$.
In Case 3, $B_{1,t}\cap B_{2,t}$ is the union of two spherical segments $S_1$ and $S_2$ in $\R^{m-1}$ with disjoint interiors. More specifically, let us identify $\R^{m-1}$ with $\R\times\R^{m-2}$ and assume, wlog, that the balls $B_{1,t}$ and $B_{2,t}$ are centered at the respective points $(0,0)$ and $(d_u,0)=(d\sqrt{1-t},0)$ in $\R\times\R^{m-2}$. Then, for $m\ge3$,
\begin{equation*}
B_{1,t}\cap B_{2,t} \\
=\{(z,y)\in\R\times\R^{m-2}\colon
|y|^2<\min[r_1^2-z^2,r_2^2-(d\sqrt{1-t}-z)^2] \\
=S_1\cup S_2,
\end{equation*}
where
\begin{equation*}
S_1:=\{(z,y)\in\R\times\R^{m-2}\colon z_1<z<r_1,|y|^2<r_1^2-z^2\},
\end{equation*}
\begin{equation*}
S_2:=\{(z,y)\in\R\times\R^{m-2}\colon d\sqrt{1-t}-r_2<z\le z_1,|y|^2<r_2^2-(d\sqrt{1-t}-z)^2\},
\end{equation*}
\begin{equation*}
z_1:=z_1(r_1,r_2)(t):=\frac{d^2 (1-t)+r_1^2-r_2^2}{2 d \sqrt{1-t}}.
\end{equation*}
Using the reflection $z\leftrightarrow d \sqrt{1-t}-z$, we see that spherical segment $S_2$ is congruent to
\begin{equation*}
S'_2:=\{(z,y)\in\R\times\R^{m-2}\colon z_2\le z<r_2,|y|^2<r_2^2-z^2\},
\end{equation*}
where
\begin{equation*}
z_2:=d \sqrt{1-t}-z_1=\frac{d^2 (1-t)-r_1^2+r_2^2}{2 d \sqrt{1-t}}
=z_1(r_2,r_1)(t).
\end{equation*}
So, in Case 3,
\begin{equation*}
g(t)=L_{m-1}(B_{1,t}\cap B_{2,t})=L_{m-1}(S_1)+L_{m-1}(S'_2) \\
=V_{r_1,r_2}(t)+V_{r_2,r_1}(t) \tag{g: Case 3}\label{cs3}
\end{equation*}
for $m\ge3$, where
\begin{equation*}
V_{r_1,r_2}(t):=\om_{m-2}\int_{z_1(r_1,r_2)(t)}^{r_1} dz\,(r_1^2-z^2)^{(m-2)/2}
=\om_{m-2}r_1^{m-1} I_{r_1,r_2}(t),
\end{equation*}
\begin{equation*}
I_{r_1,r_2}(t):=\int_{c_1(r_1,r_2)(t)}^1 dc\,(1-c^2)^{m/2-1},
\end{equation*}
\begin{equation*}
c_1(r_1,r_2)(t):=\frac{z_1(r_1,r_2)(t)}{r_1}
=\frac{d^2 (1-t)+r_1^2-r_2^2}{2 r_1 d \sqrt{1-t}} \\
=\frac{d}{2 r_1}\,(1-t)^{1/2}
+\frac{r_1^2-r_2^2}{2 r_1 d }\,(1-t)^{-1/2}. \tag{40}\label{40}
\end{equation*}
One can similarly see that \eqref{cs3} holds for $m=2$ as well.
Collecting \eqref{10}, the descriptions of Cases 1--3, \eqref{cs1}, \eqref{cs2}, and \eqref{cs3}, we get the following expression for the étendue measure of $C$:
\begin{equation*}
E=b_m\,(J_2+J_{3;1,2}+J_{3;2,1})
\tag{!}\label{!}
\end{equation*}
where
\begin{equation*}
J_2:=\om_{m-1}r_2^{m-1}\int_{t_+}^1 dt\, t^{-1/2}(1-t)^{m/2-3/2},
\end{equation*}
\begin{equation*}
J_{3;i,j}:=\om_{m-2}r_i^{m-1}\int_{t_-}^{t_+} dt\, t^{-1/2}(1-t)^{m/2-3/2}\,
\int_{c_1(r_i,r_j)(t)}^1 dc\,(1-c^2)^{m/2-1},
\end{equation*}
$b_m$ is given by \eqref{20}, $\om_k$ is given by \eqref{30}, $t_-$ and $t_+$ are defined in the descriptions of Case 1 and 2, and $c_1(r_1,r_2)(t)$ is defined in \eqref{40}.
Using the binomial expansions of $(1-c^2)^{m/2-1}$, of the latter expression for $c_1(r_1,r_2)(t)$ in \eqref{40}, and of $(1-t)^p$, one can further express the expression \eqref{!} of the étendue measure $E$ of $C$ as the ordinary integral of $dt\,t^{-1/2}$ times a triple hypergeometric-like series.
If $m$ is even, this triple series is a finite sum, and then the integral expressing $E$ is an elementary function.
In particular, for $m=2$ we get the same result as the one found in the other answer.
For $m=4$, we get
\begin{align*}
&E \\
&= \frac{4}{3} \pi ^2 \Big((r_1^3+r_2^3) \sin^{-1}(\frac{\sqrt{d^2-(r_1-r_2){}^2}}{d}) \\
&-(r_1^3+r_2^3) \sin ^{-1}(\frac{\sqrt{d^2-(r_1+r_2){}^2}}{d})
+2r_2^3 \sin ^{-1}(\frac{r_1-r_2}{d})\Big) \\
&-\frac{4 \pi ^2\sqrt{d^2-(r_1+r_2){}^2} (2 d^4-2 d^2 (7 r_1^2-r_2 r_1+7
r_2^2)-3 (r_1+r_2){}^2 (r_1^2-3 r_2
r_1+r_2^2))}{45 d^2} \\
&+\frac{4 \pi ^2\sqrt{d^2-(r_1-r_2){}^2} (2 d^4-2 d^2 (7 r_1^2+r_2 r_1+7
r_2^2)-3 (r_1-r_2){}^2 (r_1^2+3 r_2r_1+r_2^2))}{45 d^2}.
\end{align*}
