Why isn't my calculation that we should be able to see the sun well beyond the observable universe valid?

The problem with your derivation is that you distributed the photons over a 360° circle, so the photons only spread out in a two-dimensional circle. This means that the intensity of light drops off at a rate proportional to $1/r$ instead of $1/r^2$ (where $r$ is the distance from the center of the sun) like it does in a three-dimensional universe.

So, starting with $N$ photons emitted per second, the intensity of photons at a distance $r$ from the sun is given by $$I = \frac{N}{4\pi r^2}.$$ This comes from spreading out the photons over the surface of a sphere surrounding the sun.

The number of photons seen by your eye per second is just the intensity multiplied by the area of the iris of your eye: $$n = IA_\text{eye} = \frac{N}{4\pi r^2}A_\text{eye}.$$ You are looking for the distance beyond which you would see less than one photon per second: $$n = \frac{N}{4\pi r^2}A_\text{eye} \lt 1$$ Solving for $r$ gives $$r > \sqrt\frac{NA_\text{eye}}{4\pi}$$ Plugging in your numbers gives $$r > \sqrt{\frac{(10^{45})\pi(0.005\,\textrm{m}/2)^2}{4\pi}} = 4\cdot10^{19} \,\textrm{m} \approx 4000\,\textrm{light-years}$$ This distance is still well within our own galaxy.

There is a second error. Not only you distributed the photons over a circle, not a sphere, you didn't take time and cosmic expansion into account. The two are both completely separate, but both of them will each affect what you'd see.

Imagine a photon leaving the sun. Imagine somehow, your eye could see single photons, so that we don't have to worry about distance and eye sensitivity (after all, telescopes can see photons across billions of.light years. Just naked eyes can't).

Now, if your eye is cosmically speaking, near the sun - say, in the same cluster or supercluster of.galaxies - its easy. Photon arrives, you see it, done.

But you're asking about the observable universe. So the photon travels billions of years to the eye... and during that time, space itself expands. As space expands, the photons wavelength grows too. Wavelength and frequency are related, so the frequency changes. In astronomer-speak, the light is red shifted, meaning it now is longer wavelength than before (in visible light, red is the longest visible wavelength, hence the name).

The end result is, by the time photons from the suns light have travelled a chunk of the way across the visible universe, the light may well not be visible any more. It may have been stretched to infra-red or radio wavelengths instead.

Which means that even with eyes able to detect single photons, you couldn't see the sun from across the visible universe. "Detect" yes, "see" no. Because by the time its photons could reach you, they'd be outside the limits (range of visible frequencies) that the eye can actually see.

If the sun emits $N=10^{45} $ photons per second, an observer at distance $R$ will receive $N/4\pi R^2$ photons per second per unit area. If human eye has radius about $r=10^{-3} m$, it will receive 1 photon per second if

$$\frac{N\pi r^2}{4\pi R^2} =1, $$

$$R \sim \sqrt{N} r \sim 10^{20} m\sim 1 kpc. $$

This is less than the size of the universe.