This is to address one of the comments in how to derive the formula for the dihedral angles.
Given a tip of a tetrahedron, label the 3 edges attached to it as $1, 2, 3$ in such a way when you view the tip within the tetrahedron, the edges $1, 2, 3$ are arranged in counterclockwise manner.
Let $\vec{v}_1$, $\vec{v}_2$, $\vec{v}_3$ be the 3 unit vectors pointing away from the tip along the direction of edge $1, 2, 3$ respectively. Let us look at one of edge, say edge $2$. It is in contact with two faces.
One bounded by vectors $\vec{v}_1$ and $\vec{v}_2$. Another bounded by vectors
$\vec{v}_2$ and $\vec{v}_3$. The inward normal vectors of these two faces are given by:
$$
\vec{n}_{12} = \frac{\vec{v}_1 \times \vec{v}_2}{|\vec{v}_1 \times \vec{v}_2|}
\,\,\,\text{ and }\,\,\,
\vec{n}_{23} = \frac{\vec{v}_2 \times \vec{v}_3}{|\vec{v}_2 \times \vec{v}_3|}
$$
In terms of the normal vectors, the dihedral angle at edge $2$, $\phi_2$, satisfies:
$$\begin{align}
\cos(\phi_2) &= -\vec{n}_{12}\cdot\vec{n}_{23}\\
&= -\frac{( \vec{v}_1 \times \vec{v}_2 )\cdot(\vec{v}_2 \times \vec{v}_3)}{
|\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\\
&= -\frac{(\vec{v}_1\cdot\vec{v}_2)(\vec{v}_2\cdot\vec{v}_3) - (\vec{v}_1\cdot\vec{v}_3)|\vec{v}_2|^2}{|\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\\
&= \frac{\vec{v}_1\cdot\vec{v}_3 - (\vec{v}_1\cdot\vec{v}_2)(\vec{v}_2\cdot\vec{v}_3)}{|\vec{v}_1 \times \vec{v}_2||\vec{v}_2 \times \vec{v}_3|}\tag{*}
\end{align}$$
Let $\theta_{ij}$ be the vertex angle between edge $i$ and $j$. We can simplify R.H.S of (*) to get:
$$\cos(\phi_2) = \frac{\cos(\theta_{13}) - \cos(\theta_{12})\cos(\theta_{23})}{\sin(\theta_{12})\sin(\theta_{23})}$$
The formula of other dihedral angles can be derived in same manner.
The answer comes from spherical trigonometry, but don't let the "spherical" part throw you.
Let $a$, $b$, $c$ be the "face-corner" angles (bounded by pairs of edges), and let $A$, $B$, $C$ be the "dihedral angles" (bounded by pairs of faces), with $a$ opposite $A$, $b$ opposite $B$, and $c$ opposite $C$.
The two Spherical Laws of Cosines relates these angles thusly:
$$\begin{align}
\cos c &= \phantom{-}\cos a\cos b+\sin a \sin b \cos C \tag{1}\\[4pt]
\cos C &= -\cos A\cos B+\sin A\sin B \cos c \tag{2}
\end{align}$$
Note: If the plane containing face-angle $c$ is perpendicular to the edge along dihedral angle $C$, then $a=b=90^\circ$ so that $(1)$ reduces to $\cos c=\cos C$, so that $c=C$. This is consistent with OP's parenthetical about those angles matching in this situation.
FYI: There's also the Spherical Law of Sines:
$$\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C} \tag{3}$$
Best Answer
I realize that this post is years old, but the only documentation of an answer that I know of is younger. There is a formula. For dihedral angles $\theta_{ij}$ and solid angle $\Omega_i$,
$$\Omega_i = \theta_{ij} + \theta_{ik} + \theta_{il} - \pi$$ (Lemma 1).
If you have the six edge lengths of the tetrahedron, you can obtain the dihedral angles. For edges $e_{ij}$ with lengths denoted $d_{ij}$ and face area $F_l$ (the face opposite vertex $V_l$), the dihedral angle $\theta_{ij}$ is given by
$$\cos \theta_{ij}=\frac{D_{ij}}{\sqrt{D_{ijk}D_{ijl}}},$$
where
$$D_{ij}=-d_{ij}^4+ (d_{ik}^2+d_{il}^2+d_{jk}^2+d_{jl}^2-2d_{kl}^2)d_{ij}^2 +(d_{ik}^2-d_{jk}^2)(d_{jl}^2-d_{il}^2)$$
and
$$D_{ijk}=-16{F_l}^2$$
(Theorem 1).
Wirth and Dreiding (2014), Relations between edge lengths, dihedral and solid angles in tetrahedra