This answer try to derive the formula from first principle.
For an alternative derivation of the formula using relation between mean-width and integral over mean curvature, see the second half of my answer
to a related question.
Please note that the formula can fail for non-convex polyhedron
(e.g. $[0,1]^3 \setminus [\epsilon,1-\epsilon]^2\times [\epsilon,1]$)
We will assume the polyhedron at hand is a convex one.
Let $\pi_1, \pi_2$ be the projections: $
\;\;\mathbb{R}^3 \ni (x,y,z)
\;\; \overset{\pi_2}{\longmapsto}\;\;
\overset{\in \mathbb{R}^2}{(x,y)}
\;\; \overset{\pi_1}{\longmapsto}\;\;
x \in \mathbb{R}
$
Let $\mu O$ be the length/area/volume for $1/2/3$-dim geometric object $O$ in $\mathbb{R}^3$.
Let $R_3$ be a random rotation uniformly sampled over $SO(3)$.
For any convex polyhedron $P$, let $\;\nu P = \verb/E/_{R_3}[\mu \pi_1\pi_2 R_3 P]\;$ be its average caliper diameter.
Let $E(P)$ be the edges of $P$ and $E'(P) = \{ e \in E(P) : \pi_2 e \subset \partial \pi_2 P \}$. When none of the faces of $P$ is tangent to the $z$-direction, edges in $E'(P)$ will get mapped to edges of $\pi_2 P$. Furthermore, the correspondence will be bijective. In the illustration below, $P$ is a dodecahedron and $\pi_2 P$ is the gray polygon on $xy$-plane. $E'(P)$ are those red edges on $P$ and under $\pi_2$, they get mapped to the blue edges of $\pi_2 P$.
$\hspace1in$ 
If one further project these edges down to $x$-axis using $\pi_1$ and sum over their length, one will notice the sum overs those edges in $E'(P)$ whose $\pi_2 e$ is facing upwards or exactly right equals to the caliper diameter of $P$ along $x$-direction.
The same thing happens to the sum over those edges whose $\pi_2 e$ is facing downwards or exactly left. This leads to
$$\mu \pi_1\pi_2 P =
\frac12 \sum_{e \in E'(P)} \mu \pi_1 \pi_2 e
= \frac12 \sum_{e \in E(P)} \verb/1/_{E'(P)}(e) \mu \pi_1 \pi_2 e
$$
where $\displaystyle\;\verb/1/_{E'(P)}(e) = \begin{cases} 1,& e \in E'(P)\\
0, & \text{ otherwise }\end{cases}\;$
is the indicator function for $e \in E'(P)$.
Replace $P$ by $R_3 P$ and taking expectation value. Notice the probability for $R_3P$ to have any face tangent to $z$-direction is zero, one obtain
$$\nu P = \frac12 \sum_{e\in E(P)} \verb/E/_{R_3}[
\verb/1/_{E'(R_3P)}(R_3e) \mu \pi_1\pi_2 R_3 e
]
$$
For any edge $e$, let $L_e = \mu e$ be its length. Let $t_e$ be a unit vector pointing from one end-point of $e$ to another. Let $n_{e,1}$, $n_{e,2}$ be the outward pointing unit normal vectors for the two faces attached to edge $e$. Let $\psi_e = \pi - \delta_e$ be the angle between $n_{e,1}$ and $n_{e,2}$.
If one pick a unit vector $t$ and look at those $R \in SO(3)$ which make
$t_{Re} = t$. One will notice $n_{Re,1}$ and $n_{Re,2}$
lies on a circle which lives on a plane normal to $t$.
As long as $t$ is not parallel to the $z$-direction, $Re \in E'(RP)$ when the arc between $n_{Re,1}$ and $n_{Re,2}$ cut the $xy$-plane. This leads to
$$\verb/E/_{R_3}[ \verb/1/_{E'(R_3P)}(R_3e) | t_{R_3e} = t ] = \verb/Pr/_{R_3} [ R_3 e \in E'(R_3P) | t_{R_3 e} = t ] = \frac{\psi_e}{\pi} = 1 - \frac{\delta_e}{\pi} \quad $$
Aside from events of probability zero, this conditional expectation is independent of $t$, one find
$$\verb/E/_{R_3}[
\verb/1/_{R_3P}(R_3e) \mu \pi_1\pi_2 R_3 e]
= \left(1 - \frac{\delta_e}{\pi}\right) \verb/E/_{R_3}[ \mu \pi_1\pi_2 R_3 e ]
= \left(1 - \frac{\delta_e}{\pi}\right) \frac{L_e}{2}
$$
As a result,
$$\nu P = \sum_{e \in E(P)} \frac{\pi - \delta_e}{4\pi} L_e$$
