Area of Projection of Tetrahedron

As we are concerned with a ratio of areas, we can assume that the tetrahedron is inscribed in the cube $(\pm1,\pm1,\pm1)$ with vertices

$$B(1,1,1), \ A(-1,-1,1), \ C(1,-1,-1), \ D(-1,1,-1)$$

We have: $\vec{AB} \perp \vec{CD}$ (resp. the upper edge and lower edge). Their midpoints are $E(0,0,1)$, resp $F(0,0,-1)$ A rotation with angle $\alpha$ gives rise to a projection onto the following isosceles trapezoid:

where the projected points onto the plane are:

$$A_1(-\sqrt{2} \cos \alpha,1), \ B_1(\sqrt{2} \cos \alpha,1), \ C_1(\sqrt{2} \sin \alpha,-1), \ D(-\sqrt{2} \sin \alpha,-1)$$

Indeed, as $\tan \pi/3 = \sqrt{3}$, we must have:

$$\underbrace{\tan \angle C_1D_1A_1}_{\sqrt{3}}=\dfrac{EF}{|D_1F-A_1E|}$$

$$\dfrac{\sqrt{2}}{2}|\cos \alpha - \sin \alpha|=\frac{1}{\sqrt{3}}$$

with a particular solution:

$$\sin(\alpha-\pi/4)=\frac{1}{\sqrt{3}}$$

$$\alpha=\pi/4+\arcsin \frac{1}{\sqrt{3}}$$

From there, it is not difficult to find the ratio of areas.

Indeed, the area of the trapezoid is

$$Area \ = \ half basis \ \times \ height \ = \ 2 \sqrt{2}(\sin(\alpha)+\cos(\alpha))$$

$$ \ Area \ = \ 4\cos(\alpha - \pi/4)=4\cos(\arcsin(1/\sqrt{3}))$$

$$ \ Area \ = \ 4\sqrt{2/3}\tag{1}$$

As the lateral area of the tetrahedron is

$$8 \sqrt{3}\tag{2},$$

the final ratio is the quotient of (2) and (1):

$$\color{red}{\frac{\sqrt{2}}{6}}\approx 0.2357$$

which is below your estimation...

Here's a solution without trigonometry. Let's take a tetrahedron $ABCD$ where the distance $OO'$ between the midpoints of two skew edges is $2$, and the edges have then a length of $2\sqrt2$. The projection is an isosceles trapezoid $A'B'C'D'$, with altitude $OO'=2$ and bases $a=A'B'$, $b=C'D'$.

Projecting the figure on a plane perpendicular to $OO'$ (figure below) one realises that triangles $DD'O$ and $BB'O'$ are equal, hence $OD'^2+O'B'^2=O'B^2$, that is: $$ a^2+b^2=8. $$ Moreover, from $\angle O'B'D'=60°$ it follows $O'B'-OD'=OO'/\sqrt3$, that is: $$ a-b={4\over\sqrt3}. $$ Combining the above equations one finds: $$ a={2\over\sqrt3}(\sqrt2+1), \quad b={2\over\sqrt3}(\sqrt2-1) $$ and from there the area of the trapezoid can be found, leading to a ratio $$ {\text{area of trapezoid}\over\text{surface area of tetrahedron}} ={\sqrt2\over6}. $$