Sorry
@PeterDonis I was working on the math before the closure (actually since yesterday once
@externo finally specified what they were looking for) and so I wanted to post it anyway. It just took some time to do.
In Rindler (accelerating) coordinates the metric is $$ds^2 = -c^2 d\tau^2 =-\frac{a^2 x^2}{c^2} dt^2 + dx^2 + dy^2 + dz^2$$ where an observer at ##x_0=c^2/a## experiences a proper acceleration of ##a## and has undilated time. Time dilation is given by ##\frac{1}{\gamma} = \frac{d\tau}{dt}##, so for a clock at rest in the Rindler frame ##dx=dy=dz=0## and we have $$\frac{1}{\gamma}=\frac{d\tau}{dt}=\sqrt{\frac{a^2 x^2}{c^4} - \frac{dx^2+dy^2+dz^2}{c^2 dt^2}}= \frac{a x}{c^2}$$ For a clock at ##x=x_0+h## compared to a clock at ##x=x_0## this gives a time dilation ratio of $$ 1+\frac{ah}{c^2}$$ where ##ah## is the pseudo-gravitational potential in the Rindler frame. Note that ##h=0## corresponds to the reference clock which is undilated and has ##1/\gamma=1##. So ##h## functions as a parameter that identifies specific members of the Rindler congruence, e.g. a clock at ##h=0## or a clock at ##h=5 \mathrm{\ m}##.
Note that this is not a first-order approximation, but is the actual time dilation in the accelerating Rindler frame. By the equivalence principle this is the relationship that holds locally in a real gravitational field over a local region sufficiently small that tidal gravity effects are negligible.
Now, we will transform from the accelerating Rindler frame to an inertial Minkowski frame. If ##t=0## is the moment that the rocket is momentarily at rest in the aether frame, then this transforms to the aether frame, so this is the theoretical LET derivation even if we find it difficult to determine the correct ##t## as a practical matter.
In the inertial frame the worldline of the clock at ##h## is given by $$(T,X,Y,Z)=\left(\left(\frac{c}{a}+\frac{h}{c}\right) \sinh\left( \frac{at}{c} \right),\left(\frac{c^2}{a}+h\right) \cosh\left( \frac{at}{c} \right),0,0\right)$$$$ = \left(T, \left( \frac{c^2}{a}+h \right) \sqrt{1+\frac{a^2 c^2 T^2}{\left( c^2 + a h\right)^2}} ,0,0 \right)$$ where ##T## is time in the inertial frame and ##t## is time in the Rindler frame.
The metric in the inertial frame is ##ds^2=-c^2 dT^2 + dX^2 + dY^2 + dZ^2##. Calculating the time dilation for this case (here ##dX\ne 0##) we get $$\frac{1}{\gamma}=\sqrt{\frac{(c^2+ah)^2}{(c^2+ah)^2+a^2 c^2 T^2}}$$
Now, we have the time dilation in the inertial frame at any time ##T## for the Rindler clock at ##h##. This time dilation, as expected, depends on both ##T## and on ##h##. As ##T## increases the clock gets slower, regardless of ##h##. So, we can take the ratio of the time dilation for a clock at ##h=0## and a clock at some other ##h##. When we do that we get $$\left(c+\frac{ah}{c}\right)\sqrt{\frac{c^2+a^2 T^2}{(c^2+ah)^2+a^2 c^2 T^2}}$$
Now, the issue is that this is quantitatively the wrong value. Importantly, at ##T=0## there is no time dilation with this approach and furthermore the "time dilation" is a function of time, which is not correct for gravitational time dilation. In fact, this value is not even approximately correct. If we do just a first order approximation in ##h## we get $$1+\frac{a^3}{c^4+a^2 c^2 T^2}T^2 h + O(h^2)$$ which does not match the correct expression even to first order. Thus this expression is conclusively ruled out by experiment. The primary reason this is wrong is the one identified and explained to you by
@Ibix