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.
Earendel is not visible to the human eye, not even in principle
Not theoretically
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.
Not observationally
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).

What if we pretend it's visible?
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.
Flux density
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 $10^{-32}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$.
Flux
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
$$
F = f_\nu\Delta\nu = 9.6\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}.
$$
Power
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
$$
P = F A_\mathrm{eye}=4.8\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}.
$$
Number of photons
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
$$
\frac{dn}{dt} = \frac{P}{E_\mathrm{ph}} = 1.3\times10^{-5}\,\mathrm{s}^{-1}.
$$
Waiting time
Thus, the time you have to wait between each photon will, on average, be
$$
dt = \frac{1}{dn/dt}\sim1\,\mathrm{day}.
$$