Probability without lensing
The probability of any one photon of an ensemble of isotropically emitted photons is indeed proportional to ratio between the area of the detector, and the area of the "surface" of a sphere centered on the source. Note, however, that a galaxy seen 13 billion years (Gyr) back in time does not lie 13 billion lightyears (Glyr) away, because the Universe has expanded while the photon was traveling. When NASA says "13 Glyr away", it is an (in my opinion misunderstood) attempt not to confuse people.
The correct distance requires a model of the expansion. For the sake of this exercise, let's assume a galaxy seen 13 Gyr back in time. Such a galaxy has been redshifted by $z\simeq6.8$, and its current distance is $d=28.5\,\mathrm{Glyr}$ (!). If the galaxy weren't gravitationally lensed, the probability would then be
$$
\begin{array}{rcl}
P_\mathrm{no\,lens} & = & \frac{A_\mathrm{JWST}}{4\pi d^2} \\
& \simeq & \frac{25.4\,\mathrm{m}^2}{9\times10^{53}\,\mathrm{m}^2} \\
& \simeq & 3\times10^{-53}.
\end{array}
$$
Magnification factor
The probability for a lensed galaxy is equal to the "normal" probability, multiplied by the magnification factor $\mu$ which is the factor by which the measured flux is increased. This factor is typically of order a few, or a few tens. Large magnifications are of the order 100, and in the case of Earendel, the magnification is up to 100,000 (depending on which model you trust most).
So, if you consider a galaxy that has been lensed by a factor $\mu=10$, then the probability of any of the photons emitted by that galaxy hitting JWST is of the order of
$$
P_\mathrm{lens} = \mu P_\mathrm{no\,lens}\simeq 3\times10^{-52}.
$$
How many photons does JWST catch?
That's a very small number, but luckily galaxies emit many, many photons every second. A bright galaxy in these epochs would have a luminosity of, say, $L_\mathrm{gal} = 10^{10}$ times that of our Sun, or $4\times10^{36}\,\mathrm{W}$. For an order-of-magnitude estimate, let's assume that most is emitted in the optical, with an average wavelength of $\lambda=600\,\mathrm{nm}$, corresponding to an energy of $E_\mathrm{ph}=h\nu=hc/\lambda\simeq3\times10^{-19}\,\mathrm{J}$.
Hence, each second the galaxy emits
$$
\dot{n}_\mathrm{ph} = \frac{L_\mathrm{gal}}{E_\mathrm{ph}} \simeq 10^{55}\,\mathrm{photons\,per\,second}.
$$
Because of the expansion of the Universe, the rate at which we receive individual photons here is a factor of $(1+z)$ smaller than the rate at which they're emitted (thanks @jawheele for catching this).
Hence, in this case with $\mu=10$, $z=6.8$, and $L_\mathrm{gal}=10^{10}\,L_\odot$, JWST would catch
$$
n_\mathrm{ph} = \frac{\dot{n}_\mathrm{ph}}{1+z} P_\mathrm{lens} \sim 400\,\mathrm{photons\,per\,second}.
$$
Note that galaxies do not in general emit photons isotropically. In the case of dusty disk galaxies, they may emit quite a lot more in the face-on direction than in the edge-on direction.
