Assuming your eye is part of the optic system with the telescope, and assuming the exit pupil of the telescope's eyepiece isn't larger than your eye's iris (that is, your eye doesn't contribute as a secondary aperture stop), then a telescope+eye system is no different than any other imaging system. \$\require{cancel}\$
As it turns out, with some back-of-the envelope calculations, you can show that for any (lossless) optical system that produces a photographic image of the sun (that is, we're not talking about solar concentrators for solar thermal power), the intensity of the sun's image on the sensor is approximately $${12\over N^2}\,{\mathrm{W}\over\mathrm{mm^2}}$$
That is, for any square-millimeter section of the image sensor (or your eye's retina) that is completely covered by part of the sun's image (i.e., inside the sun's disk), that area is receiving thermal energy of about 12 watts divided by the f-number of the lens/telescope squared. That's a lot of thermal energy for your retina, and for most imaging sensors. Some cameras sensors can handle that intensity for a short period of time because the actual image area is so small (such as with smaller phone cameras, or for very wide-angle lenses that produce such a small sun image).
The maths of the above value are fairly simple. Rectilinear lenses (any lens used to image a solar eclipse will be rectilinear, because you're interested in apparent magnification) obey the pinhole projection formula, which is simply that rays entering the lens at an angle \$\alpha\$ to the optical axis exit at the same angle. This implies that a subject of diameter \$d_\mathrm{s}\$ in focus at a distance \$u\$ from the camera will result in a image of diameter \$d_\mathrm{i}\$ (with a rear focus distance of \$v\$) according to similar triangles:
$$
{d_\mathrm{s} \over u} = {d_\mathrm{i} \over v}
$$
For subjects very far away (such as the astronomical objects), the rear focus distance is essentially identical to the lens's focal length \$f\$: \$v \approx f\$. Thus, astronomical objects of diameter \$d_\mathrm{s}\$ at a distance \$u\$ from the Earth will result in a focused image of diameter \$d_\mathrm{i} = f\cdot (d_\mathrm{s}/u)\$.
The ratio of the Earth–Sun distance (\$1\,\mathrm{AU}\$) to the sun's diameter \$d_☉\$ is 107.5, or about \$108:1\$. So the diameter of the image of the sun on the camera's sensor is
$$
d_\mathrm{i} = {f \over 108}\,.
$$
Note that this is regardless of the lens's diameter, the lens's aperture size, or the size of the lens's entrance pupil.
For example, capturing the sun during an eclipse with a 500 mm lens will result in an image of \$500\,\mathrm{mm}/108 = 4.6\,\mathrm{mm}\$ on the sensor. Note that this is just the disc of the sun. This does not include the coronal halo, which is substantially larger, and usually of interest during an eclipse.
Roughly speaking, under direct sunlight on a clear day, the incoming solar radiation (insolation), the amount of power per area, is nominally \$\mathrm{I_s} = 1\,{\mathrm{kW/m^2}}\$. So aiming your camera directly at the sun, the lens's entrance pupil has a diameter \$D = f/N\$, and therefore experiences the power of \$\mathrm{I_s}\$ times the area of the entrance pupil:
$$
\begin{align*}
P &= \mathrm{I_s} \times \text{entrance pupil area} \\
&= \mathrm{I_s} \times \pi\;\left({D\over2}\right)^2 \\
&= \mathrm{I_s} \times {\pi\over4}\left({f\over N}\right)^2 \\
\end{align*}
$$
All of that incoming power is focused onto the sun's image on the sensor. We already know the sun's image has a diameter of \$d_\mathrm{i}=f/108\$. So in terms of power per unit area of the sun's image on sensor, the sun-image sensor area experiences an irradiance of that power \$P\$ divided by the sun's image area (\$A_\mathrm{i} = \pi(\tfrac{d_\mathrm{i}}{2})^2\$):
$$
\begin{align*}
I &= {P \over \text{sun image area}} \\
&= P \times {1\over \pi} \left({2\over d_\mathrm{i}}\right)^2 \\
&= P \times {4\over\pi}\left({108\over f}\right)^2 \\
&= \mathrm{I_s}\times \bcancel{\pi\over 4}{\cancel{f^2}\over N^2}\bcancel{4\over\pi}{108^2\over \cancel{f^2}} \\
&= {1\,\mathrm{kW}\over \mathrm{m^2}} \cdot {108^2\over N^2} \\
&\approx {12\over N^2}\,{\mathrm{MW\over m^2}} = {12\over N^2}\,{\mathrm{W\over mm^2}}
\end{align*}
$$
This feels like a surprising fact, that regardless of the lens size, camera size, etc., the power intensity per unit area of the imaged sun on the sensor is only affected by the lens's aperture number. So a big DSLR with a fast \$f/2\$ lens results in \$3\,\mathrm{W/mm^2}\$, whereas an iPhone's \$f/1.78\$ lens results in 27% higher radiant power intensity, at \$3.8\,\mathrm{W/mm^2}\$. Of course, the difference is that the DSLR is producing that power intensity over a larger area on the sensor than the iPhone is. And assuming that the radiant heat distribution is approximately linear through the thin plane of the image sensor, the DSLR suffers from a larger Area/Circumference ratio of the sun's image on sensor. Thus, the thermal resistance to radiant heat dissipation issue scales, unfavorably, linearly with the lens's focal length.
