For a Schwarzschild observer the gravitational time dilation means the local time at a distance $r$ from a mass $M$, appears to be running slow by a factor of:
$$ \sqrt {1 - \frac{2GM}{r c^2}} $$
You get this straight from the Schwarzschild metric:
$$ ds^2 = -\left(1-\frac{2GM}{r c^2}\right)dt^2 + \left(1-\frac{2GM}{r c^2}\right)^{-1}dr^2 + r^2 d\Omega^2 $$
because hovering above the Sun, both $dr$ and $d\Omega$ are zero.
Anyhow, I reckon the acceleration due to the Sun's gravity is 9.81m/s$^2$ at a distance of about $3.68 \times 10^9$m (using the Newtonian expression for $a$), and at this distance I work out the time dilation to be about 12.6 seconds per year. So the time dilation is not simply proportional to the gravitational acceleration.
You can get an approximate, but very good, relationship between the acceleration and the time dilation by using the Newtonian expression for the acceleration to get $r$ and substituting in the expression for the time dilation. Doing this gives:
$$ \sqrt{1 - \frac{2\sqrt{aGM}}{c^2}} $$
which I don't think is terribly illuminating!
Special relativity (let's leave aside GR for now) is notoriously unintuitive - generations of physics students have found this to their cost, so you are far from alone. So there is no simple intuitively clear explanation over what is going one. That said, I will attempt a quasi-intuitive explanation.
I think the mistake students make is to take time dilation in isolation. It's easy to think here's a phenomenon called time dilation: what causes it? What actually happens is that different observers will disagree about what constitutes space and what constitutes time and time dilation is just part of a bigger phenomenon.
As I sit here at my keyboard I'm not moving in space, but I am moving in time. So for some activity (e.g. from the start to the end of me typing this sentence) my $\Delta x = 0$ and my $\Delta t = T$ for some time $T$. However the bug eyed alien that has just zoomed past at $0.9c$ disagrees. The alien, seated at their own typewriter, sees me moving at $-0.9c$, so in between starting and finishing my sentence the alien sees that I have moved some distance $\Delta x = d$. But in SR space and time are linked, so if the alien measures a different $\Delta x$ they must also measure a different $\Delta t$. The two are linked by the relationship:
$$ c^2 \Delta t^2 - \Delta x^2 = c^2 \Delta t'^2 - \Delta x'^2 $$
where the unprimed $t$ and $x$ are what I measure and the primed $t'$ and $x'$ are what the alien measures. Even without doing the maths it should be obvious that because $\Delta x < \Delta x'$ it follows that $\Delta t < \Delta t'$. In other words when the alien times how long it takes me to type the sentence they measure a longer time than I do. For the alien my time has been dilated.
You've probably also heard of length contraction. Well this is the other side of the coin. Time dilation and length contraction always occur together because in effect some of the time is being converted into length and vice versa.
All this follows from a fundamntal symmetry called Lorentz invariance. This states that if we measure a property called the line interval and defined by:
$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$
then the quantity $ds^2$ is an invariant and all observers will measure the same value for it. To get the equation above linking my $(t, x)$ with the alien's $(t', x')$ I just exploited this invariance to require that $ds^2 = ds'^2$.
All very well, but I have really only pushed the non-intuitiveness one stage farther down, since my explanation assumes Lorentz invariance and this in turn is unintuitive. Still, hopefully this allows you to get some handle on what is going on.
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There is in principle a "gravitational blueshift" for signals traveling from Voyager to us. The data rate we receive will be higher than the data rate transmitted by a factor $(1+\Delta\Phi/c^2)$, where $\Delta\Phi$ is the difference in Newtonian gravitational potential between the locations.
(Of course, this formula is only valid in weak gravitational fields, where it makes sense to talk in terms of Newtonian gravitational potentials.)
If I'm not mistaken, $\Delta\Phi/c^2\approx 10^{-8}$ for this sort of system, so the shift is quite small in practice. In particular, it's much smaller than than the ordinary Doppler effect due to the fact that both Voyager and Earth are moving. A motion at a speed of just 3 m/s would cause a Doppler shift as large as this gravitational shift, and both bodies are moving much faster than that.