The following holds for a specific case where the volume enclosed by 3 spheres is contained within the triangle formed by the 3 sphere centers.
I am hesitant to put this answer up even though I feel it is correct, because the closed form I have found is actually quite ugly. The derivation is quite pretty but the resulting integral is ugly. I will discuss it a bit at the end.
I will post my thoughts anyways in the hope that someone can come along and simplify these formulas into a nicer form, or perhaps find a much simpler formula altogether!
Derivation
Let $T$ be the solid triangle defined by sides of length $a,b,c$. Define 3 disks of raius $r$: $D,E,F$ and let each vertex of $T$ be the center of a disk in the natural way. See figure included below (at the very bottom) for a graphic description.
Let the wedges be
$W_1 := D \cap T$ $W_2 := E \cap T$ $W_3 := F \cap T$
Let the lenses be
$L_1 := D \cap E$ $L_2 := E \cap F$ $L_3 := F \cap D$
and let the rounded triangle be
$ R := D \cap E \cap F .$
It should be noted that each $L_i$ is bisected by a side of $T$. That is
$ \left| L_i \cap T \right| = \left| L_i \cap \bar{T} \right| $ where $\bar{T}$ denotes the compliment of $T$ and $| \cdot |$ denotes area. So naturally, define the half-lenses
$H_1 := T \cap L_1 $ $H_2 := T \cap L_2 $ $H_3 := T \cap L_3 $
Let $ f(r) = \left| R \right| $, it follows that the volume $V$ contained in the overlap of 3 spheres with equal radii $r_s$ is equal to
$ V = 2 \int _0 ^{m} f\left(\sqrt{r_s^2-h^2}\right) dh$
where $m = \sqrt{r_s^2 - \left( \frac {\max\{a,b,c\}} {2} \right)^2}.$
It remains to find a closed form of $f(r)$. We can find one applying the inclusion exclusion principle as
$ f(r) = |T| - \sum_i |W_i| + \sum_i |H_i| $
We have $\sum |W_i| = \frac{\pi}{2} r^2$
since the sum of the internal angles of $T$ is $\pi.$
By Heron's elegant formula we have
$ |T| = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{1}{2} (a+b+c).$
Lastly, we can compute the area of each half-lense with
$|H_i| = r^2 \cos^{-1} \left( \frac{d_i}{2r} \right) - \frac{1}{4} d_i \sqrt{4r^2-d_i}$ where $d_i = c,b,a$ for $i=1,2,3$ using the formulas for lenses but halved.
Discussion
This seems to be a closed form, as the most difficult integrals to compute, namely those corresponding to the $|H_i|$ terms are found to have a closed form as per a quick query here and here on wolfram's alpha.
But because these indefinite integrals are so ugly, I fear this closed form has little practical use unless a simplification or other formula can be found!
Though one may still be able to find a tight upper and lower bound by simplifiying/approximating these ugly integrals.
