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Since a telescope is just a very powerful eye, if it took the light of Earendel 12.9 billion years for it to reach the Hubble telescope, how long would it take for that same light to reach a human eye? I guess the light from Earendel would need to travel even more so that I can see it.
Or is the distance so minimal that it's not even worth considering?
It's just a theoretical question, since I assume I will never see Earendel from my balcony.
Hubble is in low Earth orbit which is almost the same as being on the Earth's surface. So the time difference will be a just a few tens of milliseconds depending on where you are and where Hubble is around the curve of the Earth's surface. Yes, - definitely negligible on the scale of 10.9 billion years!
That famous photograph is a result of 9 hours of staring with a 2.4 meter (corrected per @pela coment) diameter pupil (mirror). Your pupil is ~300 times smaller than that in diameter and a 100,000 times (=3002) smaller in area, so to see the same number of photons, you would have to stare at the same point in the sky for about 100,000 times longer which is about 100 years. That photo may have as few as 100 photons correspnding to Earendel, so if you stare at the right spot in the sky for one year, your eye may eventually catch one photon that came from Earendel.
So if you're intent on seeing Earendel from your balcony you will need a very big pot of coffee to keep you awake and I'd recommend some warm clothing in case it gets chilly.
[Edit: PS. Please see @pela's excellent detailed answer. Earendel cannot be seen at all at visible wavelengths]
Roger Wood answers perfectly how the extra distance from Hubble to us is negligible, and estimates how much longer than Hubble you would need to look in order to detect a single photon with your eye. I'd just like to provide a little insight into exactly how long time you would have to wait to detect that photon.
The human eye can only detect light in the wavelength range $\lambda \simeq [400,700]\,\mathrm{nm}$, which astronomers like to call $[4000,7000]\,\mathrm{Å}$. Because Earendel is seen at a redshift of $z=6.2$, in order to be visible when the light reaches you, it must have been a factor $(1+z)$ shorter when it was emitted, i.e. in the range ~500–1000 Å. This is in the far/extreme ultraviolet. But at that epoch — when the Universe was less than a billion years old — although reionization was about to complete there was still plenty of neutral gas. Hence, virtually everything below the ionization threshold at 917 Å was absorbed close to, if not inside, Earendel's host galaxy.
Photons of longer wavelengths may escape the host, but as they're redshifted they eventually reach the wavelength 1216 Å which corresponds to the Lyman α transition between hydrogen's ground and first excited state. Neutral hydrogen thus scatters the photons out of the line of sight, effectively absorbing them from our observations. This is the so-called Gunn-Peterson trough.
Hence, from a galaxy this far, only photons that are emitted redward of 1216 Å have a fair chance of reaching us, at which point they'll be 1216 Å × (1+z) ≈ 0.9 µm or longer, which is in the infrared (IR) region, invisible to the human eye.
Indeed, all observations of Earendel taken in optical filters are consistent with having zero flux, as seen in this table from the paper presenting Earendel, Welch et al. (2022) (I marked them green).
We could pretend that humans are able to detect IR photons, or we could pretend that the emitted UV photons weren't absorbed. As the former would require us to define an arbitrary range in which we could detect the photons, let's go with the latter.
As marked in red in the table, the detected fluxes are of the order of 20–50 nJy. A "nJy", or nano-Jansky, is a measure of flux density, i.e. flux per frequency bin, and is equal to $10^{-32}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$.
Let's assume that the far-UV radiation that we're pretending not to be absorbed would have a similar flux density of $f_\nu = 30\,\mathrm{nJy}$. The visible range of [4000–7000] Å corresponds to a frequency range of $[4.3\text{–}7.5]\times10^{14}\,\mathrm{Hz}$, i.e. $\Delta\nu = 3.2\times10^{14}\,\mathrm{Hz}$. Then the total flux piercing your eye is $$ F = f_\nu\Delta\nu = 9.6\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}. $$
In the dark, your pupil expands to roughly 8 mm in diameter, so its area is $A_\mathrm{eye}\sim50\,\mathrm{mm}^{2}$. The total power entering your pupil is then $$ P = F A_\mathrm{eye}=4.8\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}. $$
The typical energy of a visible photon can be taken as that of a green photon with $\lambda\sim5500\,\mathrm{Å}$, i.e. $E_\mathrm{ph} = h\nu = hc/\lambda = 3.6\times10^{-12}\,\mathrm{erg}$. Hence, the number rate of photons entering the eye is $$ \frac{dn}{dt} = \frac{P}{E_\mathrm{ph}} = 1.3\times10^{-5}\,\mathrm{s}^{-1}. $$
Thus, the time you have to wait between each photon will, on average, be $$ dt = \frac{1}{dn/dt}\sim1\,\mathrm{day}. $$
