The volume of a tetrahedron cannot be determined from the surface areas of the faces. I shall provide a family of counterexamples.
A tetrahedron is called equifacial if its four faces are all congruent triangles. Equivalently, a tetrahedron is equifacial if its opposite edges have the same length.
Starting with any triangle in the plane, you can try to "glue" together four copies of the triangle in an obvious way to get a tetrahedron. This process has the following properties:
- If you start with an equilateral triangle, the result is a regular tetrahedron.
- If you start with an acute triangle, the result is an equifacial tetrahedron.
- If you start with an right triangle, the result is a "degenerate" tetrahedron with zero volume.
- If you start with an obtuse triangle, you can't make a tetrahedron.
Now, it is possible to continuously deform an equilateral triangle into a right triangle without changing the area. Therefore, it is possible to continuously deform a regular terahedron into a degenerate tetrahedron without changing the areas of the faces!
The following animation shows this process:
All of the tetrahedra shown in this animation have faces with area 1, but the volume decreases continuously from about $0.41$ to $0$.
Of course, equifacial tetrahedra aren't the only possible counterexample. Indeed, for any allowed quadruple $(S_1,S_2,S_3,S_4)$ of areas, there ought to be a whole interval of possible values for the volume.
Comment expanded to answer per request.
First we need a lemma
A tetrahedron with vertices $p_1, \ldots,p_4$ admits a midsphere (ie. a sphere tangent to all its edges) if and only if one can find positive numbers $r_1,\ldots, r_4 $ to represent the edge lengths in following manner: $$\ell_{ij} = |p_i - p_j| = r_i + r_j\quad\text{ for } i \ne j$$
The only if part is trivial. For the if part, since I cannot locate a proof online, I have outlined a proof at the end.
Back to the problem at hand. Start from all $r_i = 1$. If one randomly perturb them by small and distinct amounts. With probability one, one can find a set of $\ell_{ij}$ all different, realizable as edge lengths of tetrahedron. By above
lemma, above tetrahedron admits a mid-sphere.
proof of 'if' part of lemma.
Let $r_1,\ldots,r_4 > 0$ and $p_1,\ldots,p_4 \in \mathbb{R}^3$ satisfies
$|p_i - p_j | = r_i + r_j$ whenever $i \ne j$.
Given any two points $u,v$, let $(\delta_1,\cdots,\delta_4)$ be
the differences of their barycentric coordinates with respect to tetrahedron $p_1p_2p_3p_4$. It is not hard to show:
$$|u-v|^2 = G(\delta_1,\ldots,\delta_4) \stackrel{def}{=} 2 \sum_{k=1}^4 r_k^2 \delta_k^2 - \left(\sum_{i=1}^4 r_k\delta_k\right)^2\tag{*1}$$
Let $\displaystyle\;\rho_1 = \sum_{k=1}^4 \frac{1}{r_k}\;$ and $\displaystyle\;\rho_2 = \sum_{k=1}^4 \frac{1}{r_k^2}\;$.
Let $q$ be the point with barycentric coordinates $(q_1,\ldots,q_4)$ where
$\displaystyle\;q_k = \frac{1}{\rho_1^2 - 2\rho_2}\left(\frac{\rho_1}{r_k} - \frac{2}{r_k^2}\right)\;$.
For any edge $p_ip_j$, let $t_{ij}$ be the point on it with $|p_i - t_{ij}| = r_i$ and $|p_j - t_{ij}| = r_j$.
For example, consider the edge $p_1p_2$. The barycentric coordinate of $t_{12}$
is $\left(\frac{r_2}{r_1+r_2},\frac{r_1}{r_1+r_2},0,0\right)$.
Throw this into $(*1)$, we find
$$\begin{align}
|q - t_{12}|^2 &= G\left(q_1 - \frac{r_2}{r_1+r_2}, q_2 - \frac{r_1}{r_1+r_2},q_3,q_4\right) \\
&= G(q_1,q_2,q_3,q_4) + \frac{4r_1r_2}{r_1+r_2}(q_3 r_3 + q_4 r_4)
\end{align}$$
With a little bit of algebra, one can show that
$$G(q_1,q_2,q_3,q_4) = \frac{-4}{\rho_1^2 - 2\rho_2}$$
Together with
$$q_3r_3 + q_4r_4 = \frac{2}{\rho_1^2-2\rho_2}\left(\rho_1 - \frac{1}{r_3}-\frac{1}{r_4}\right) = \frac{2}{\rho_1^2 - 2\rho_2}\frac{r_1 + r_2}{r_1r_2}$$
We find
$$|q - t_{12}|^2 = r_m^2 \stackrel{def}{=} G(q_1,\ldots,q_4) + \frac{8}{\rho_1^2 - 2\rho_2}
= \frac{4}{\rho_1^2 - 2\rho_2}
\tag{*2}$$
Since $r_m$ is independent of our choice of edge $t_{ij} = t_{12}$. This means the six point $t_{ij}$ are at a common distance $r_m$ to $q$.
Notice
$$|q - p_1|^2 = G(q_1 - 1,q_2,q_3,q_4)
= G(q_1,q_2,q_3,q_4) + \underbrace{2r_1(-q_1r_1 + q_2r_2 + q_3r_3 + q_4r_4)}_{X} + r_1^2
$$
and the mess $X$ equals to $\displaystyle\;\frac{2r_1}{\rho_1^2 - 2\rho_2}\left(2\rho_1 + \frac{2}{r_1} - \frac{2}{r_2} - \frac{2}{r_3} - \frac{2}{r_4}\right) = \frac{8}{\rho_1^2 - 2\rho_2}$.
We find
$$|q-p_1|^2 = |q - t_{12}|^2 + r_1^2 = |q - t_{12}|^2 + |t_{12} - p|^2$$
This implies $q - t_{12}$ is perpendicular to $t_{12} - p_1$. By a similar argument,
we find $q - t_{ij}$ is perpendicular to $t_{ij} - p_i$ for all $i \ne j$.
As a result, the sphere centered at $q$ with radius $r_m$ is tangent to the edges
$p_1p_j$ at $t_{ij}$ and the 'if' part of lemma follows.
Upate
One may wonder what happens if $\rho_1^2 - 2\rho_2$ vanishes or even goes to negative. Let $V$ be the volume of tetrahedron, we can compute it using the Cayley-Menger determinant of the edge lengths $\ell_{ij}$:
$$288V^2 = \verb/CM/ \stackrel{def}{=}
\left|\begin{matrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & \ell_{12}^2 & \ell_{13}^2 & \ell_{14}^2\\
1 & \ell_{12}^2 & 0 & \ell_{23}^2 & \ell_{24}^2\\
1 & \ell_{13}^2 & \ell_{23}^2 & 0 & \ell_{34}^2\\
1 & \ell_{14}^2 & \ell_{24}^2 & \ell_{34}^2 & 0\\
\end{matrix}\right|$$
Substitute $\ell_{ij}$ by $r_i+r_j$ and with help of an CAS, one find
$$288 V^2 = \verb/CM/ = 32(abcd)^2(\rho_1^2 - 2\rho_2)
\quad\implies\quad \rho_1^2 - 2\rho_2 = \left(\frac{3V}{r_1r_2r_3r_4}\right)^2$$
For non-degenerate tetrahedron, $V > 0 \implies \rho_1^2 - 2\rho_2 > 0$.
When $\rho_1^2 - 2\rho_2 \to 0$, $V \to 0$. The tetrahedron becomes planar and hence the midradius $r_m$ diverges.
If one pick $r_k$ to make $\rho_1^2 - 2\rho_2 < 0$, the corresponding $\verb/CM/ < 0$ and the set of edge lengths $r_i + r_j$ cannot be realized as edge lengths of any tetrahedron.
As a side note, if one compare above results with $(*2)$, we obtain following interesting relation:
$$3Vr_m = 2r_1r_2r_3r_4$$
Best Answer
Introducing conventions familiar to me, I'll consider tetrahedron $OABC$ to have edges and face-angles about $O$, and face-areas
$$a := |OA| \qquad b := |OB| \qquad c := |OC| \\[6pt] \alpha := \angle BOC \qquad \beta := \angle COA \qquad \gamma := \angle AOB \\[6pt] W := |\triangle ABC| \qquad X := |\triangle OBC| \qquad Y := |\triangle OCA| \qquad Z:=|\triangle OAB|$$ Without fear of confusion, I'll also use $A$, $B$, $C$ to refer to the dihedral angles along edges $a$, $b$, $c$.
As OP notes, insphere-determined sub-triangles that share an edge have the same area. Let $\sigma_a$, $\sigma_b$, $\sigma_c$ be the areas of the triangles sharing respective edges $a$, $b$, $c$; further, let $\sigma_d$, $\sigma_e$, $\sigma_f$ be the areas of the triangles sharing edges $|BC|$, $|CA|$, $|AB|$. (I typically denote those edges $d$, $e$, $f$, but that's not important here.) Of course, the full face-areas are simple sums of these sub-areas:
$$W = \sigma_d + \sigma_e + \sigma_f \qquad X = \sigma_d+\sigma_b+\sigma_c \qquad \text{etc} \tag1$$
Below I show that $\cos A$, $\cos B$, $\cos C$ are also expressible in terms of the $\sigma$s, which in turn guarantees a unique volume, via $$81V^4 = 4X^2Y^2Z^2(1-2\cos A\cos B\cos C-\cos^2 A-\cos^2 B-\cos^2 C) \tag2$$
To get at the cosine relations, let's coordinatize: $$O=(0,0,0) \qquad A = (a,0,0) \qquad B = (b \cos\alpha,b\sin\alpha, 0) \\[6pt] C = (c\cos\beta,c\sin\beta\cos A,c\sin\beta\sin A)$$
It is "known" that the incenter of $OABC$ is given by $$I = \frac{W O + X A + Y B + Z C}{W+X+Y+Z} \tag3$$
If $P$ is the point where the insphere touches face $Z$ (the $xy$-plane), then $$\begin{align} \sigma_a := |\triangle OAP| &= \frac12|OA|P_y = \frac12 a \frac{Yb\sin\gamma+ Z c \sin\beta \cos A}{W+X+Y+Z} \\[8pt] &= \frac{YZ(1+\cos A)}{W+X+Y+Z} \tag4 \end{align}$$ Likewise, $$\sigma_b = \frac{ZX(1+\cos B)}{W+X+Y+Z} \qquad \sigma_c = \frac{X Y(1+\cos C)}{W+X+Y+Z} \tag5$$
Substituting $X$, $Y$, $Z$, $\cos A$, $\cos B$, $\cos C$ in terms of the $\sigma$s into $(2)$, we find
As a sanity check, a regular tetrahedron with side-length $s$ has equilateral faces of area $W=X=Y=Z=\frac14s^2\sqrt{3}$ and sub-faces of area $\sigma_a=\cdots=\sigma_f = \frac1{12}s^2\sqrt{3}=p=q=r$. As a result, $(\star)$ yields $$V^4 = \frac{16}{81} \left(6\cdot \frac1{12}s^2\sqrt{3}\right)^2\left(3\cdot \frac1{12}s^2\sqrt{3}\right)\left(\frac1{12}s^2\sqrt{3}\right)^3 = \left(\frac1{12} \sqrt{2} s^3\right)^4$$ as expected.