[Edit I've found a determinant that satisfies the letter of the previous version of this question, but not its "Cayley-Menger" spirit.]
A tetrahedron with face areas $w$, $x$, $y$, $z$ and "pseudo-face" areas $h$, $j$, $k$ has volume $V$ given by
$\begin{eqnarray} 81 V^4 &=& 2 w^2 x^2 y^2 + 2 w^2 y^2 z^2 + 2 w^2 z^2 x^2 + 2 x^2 y^2 z^2 + h^2 j^2 k^2 \\[4pt] &-&h^2(w^2 x^2+y^2 z^2)-j^2(w^2 y^2+z^2 x^2)-k^2(w^2 z^2+x^2 y^2) \end{eqnarray}$
All the cool kids in the world of tetrahedral formulas are the determinants or minors of some matrix or another, such as the Cayley-Menger or the Gram. I seek to express the above volume formula in such a form.
Note that, with $A$, $B$, $C$ the dihedral angles between respective face-pairs $(y,z)$, $(z,x)$, $(x,y)$, we have this formula for volume:
$81 V^4 = 4 x^2 y^2 z^2 \left( 1 - 2 \cos{A} \cos{B} \cos{C} - \cos^2 A - \cos^2 B - \cos^2 C \right) = 4 x^2 y^2 z^2 \left|\begin{array}{ccc} 1 & -\cos C & -\cos B \\ -\cos C & 1 & -\cos A \\ -\cos B & -\cos A & 1 \end{array}\right|$
We also have what I call the Second Law of Cosines ...
$h^2 = y^2 + z^2 - 2 y z \cos A \qquad j^2 = z^2 + x^2 - 2 z x \cos B \qquad k^2 = x^2 + y^2 - 2 x y \cos C$
... which gives rise to this determinant form of the volume formula:
$81V^4 = \frac{1}{2} \left|\begin{array}{ccc} 2 x^2 & k^2-x^2-y^2 & j^2-z^2-x^2 \\ k^2-x^2-y^2 & 2 y^2 & h^2-y^2-z^2 \\ j^2-z^2-x^2 & h^2-y^2-z^2 & 2 z^2 \end{array}\right| $
Verifying equality with my target volume formula requires the tetrahedral "Sum of Squares" identity:
$w^2+x^2+y^2+z^2=h^2+j^2+k^2$
Now, while I'm pleased to have a reasonably-straightforward determinant equation (thus answering my original question myself), that particular determinant doesn't quite seem a worthy peer of the Cayley-Menger ... perhaps because it doesn't treat all face areas equally.
Can we do better?
(Yes, "better" is subjective, but I think we know what I mean: it should have the Cayley-Menger spirit. I suppose my primary criterion is that the determinant be symmetric in the four face areas and the three pseudo-face areas; ideally, the individual entries are (subjectively) "uncomplicated".)
Here's a possibly-helpful discussion from the previous version of this question:
If $a$, $b$, $c$, $d$, $e$, $f$ are edges of the tetrahedron with $w = \triangle def$, $x = \triangle dbc$, $y = \triangle aec$, and $z = \triangle abf$, then $ 9 V^2 a^2 = [h,y,z] \qquad 9 V^2 \; b^2 = [j,z,x] \qquad 9 V^2 \; c^2 = [k,x,y]\\ 9 V^2 d^2 = [h,w,x] \qquad 9 V^2 \; e^2 = [j,w,y] \qquad 9 V^2 \; f^2 = [k,w,z] $ where "$[\bullet]$" is the "Heronic product" $\begin{eqnarray} [p,q,r] &:=& (p+q+r)(-p+q+r)(p-q+r)(p+q-r)\\ &=&-p^4-q^4-r^4+2p^2q^2+2q^2r^2+2r^2p^2 \end{eqnarray}$ Consequently, we can substitute into the Cayley-Menger determinant formula to write $\begin{eqnarray} 288 V^2 &=& \left| \begin{array}{ccccc} 0 & a^2 & b^2 & c^2 & 1 \\ a^2 & 0 & f^2 & e^2 & 1 \\ b^2 & f^2 & 0 & d^2 & 1 \\ c^2 & e^2 & d^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array}\right| \\ &=& \frac{1}{(9V^2)^3}\left| \begin{array}{ccccc} 0 & [h,y,z] & [j,z,x] & [k,x,y] & 1 \\ \; [h,y,z] & 0 & [k,w,z] & [j,w,y] & 1 \\ \; [j,z,x] & [k,w,z] & 0 & [h,w,x] & 1 \\ \; [k,x,y] & [j,w,y] & [h,w,x] & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array}\right|\\ &=:&\frac{1}{729V^6} \mathrm{det} M \end{eqnarray} $ Since $\mathrm{det}M$ is proportional to the 8th-power of $V$ ---and therefore, the square of the target formula--- perhaps I should attempt to construct $\sqrt{M}$.
My investigations of matrix $M$ have been fruitless. (Note, though, that despite its complexity ---and the fact that it generates the square of what I want--- $M$ has the kind of symmetry I would expect from a properly Cayley-Menger-esque result.) I think it's time to add a bounty to this question. [Bounty expired.]
Edit Here's a tantalizing alternative "(negative-)square-of-what-I-want" matrix:
$N := \left|\begin{array}{cccccc} 0 & z^2 & y^2 & h^2 & y^2 & z^2 \\ z^2 & 0 & x^2 & x^2 & j^2 & z^2 \\ y^2 & x^2 & 0 & x^2 & y^2 & k^2 \\ h^2 & x^2 & x^2 & 0 & w^2 & w^2 \\ y^2 & j^2 & y^2 & w^2 & 0 & w^2 \\ z^2 & z^2 & k^2 & w^2 & w^2 & 0 \end{array}\right| \qquad \mathrm{det}N = -\left(81 V^4\right)^2$
The pattern in the entries can be described thusly:
Index the rows and columns with edges $(a,b,c,d,e,f)$. Then the $pq$-th element ($p\ne q$) corresponds to the face determined by edges $p$ and $q$. (Opposite edges $p$ and $q$ determine a pseudo-face.)
The minors of $N$ are very interesting. Writing $N_p$ for the matrix obtained by removing row and column $p$ from $N$:
$\begin{array}{ccc} \mathrm{det} N_a = 2 w^2 x^2 \cdot 81 V^4 & \mathrm{det} N_b = 2 w^2 y^2 \cdot 81 V^4 & \mathrm{det} N_c = 2 w^2 z^2 \cdot 81 V^4 \\ \mathrm{det} N_d = 2 y^2 z^2 \cdot 81 V^4 & \mathrm{det} N_e = 2 z^2 x^2 \cdot 81 V^4 & \mathrm{det} N_f = 2 x^2 y^2 \cdot 81 V^4 \\ \end{array}$
Note, for instance in the case of $N_a$, that faces $w$ and $x$ meet along edge $d$, which is opposite edge $a$.
Moreover, with $N_{pq}$ the matrix obtained by removing row $p$ and column $q$ (with $p\ne q$):
$\mathrm{det} N_{ab} = - w^2 \left( k^2 - x^2 - y^2 \right) \cdot 81 V^4 = 2 w^2 x y \cos C \cdot 81 V^4 \qquad \text{etc}$
I get the sense that this gets me closer to my goal.
Edit. Getting even closer. Augmenting matrix $N$ into more Menger-like form gives
$\begin{align} P &:= \left|\begin{array}{ccccccc} 0 & z^2 & y^2 & h^2 & y^2 & z^2 & 1 \\ z^2 & 0 & x^2 & x^2 & j^2 & z^2 & 1 \\ y^2 & x^2 & 0 & x^2 & y^2 & k^2 & 1 \\ h^2 & x^2 & x^2 & 0 & w^2 & w^2 & 1 \\ y^2 & j^2 & y^2 & w^2 & 0 & w^2 & 1 \\ z^2 & z^2 & k^2 & w^2 & w^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array}\right| \\[6pt] \mathrm{det} P &= 2\left(w^4+x^4+y^4+z^4-h^2j^2-j^2k^2-k^2h^2\right) \cdot 81 V^4 \end{align}$
This may be about as good as I can expect. I'd prefer, though, that the multiplied polynomial --if there must be one-- have an unambiguous sign (and vanish only trivially).