Let's try to work out the formula for the amount of crush you'd experience if you attempted to hover over a Schwarzschild black hole to get some time dilation factor ##\gamma##.
First we need the formula for the amount of crush, which is the proper acceleration. I believe the formula is:
$$a = \frac{GM}{r^2\sqrt{1-\frac{r_s}{r}}}$$
for instance
https://en.wikipedia.org/wiki/Komar_mass which assumes that G=1, or
https://www.quora.com/What-would-the-acceleration-due-to-gravity-be-on-the-surface-of-a-black-hole. I recall seeing this formula in Wald's book "General Relativity", but not which page I saw it on, if I had more time I'd look it up to verify my fallible memory.
Gravitational time dilation is given by
$$\gamma = \frac{1}{\sqrt{1-\frac{r_s}{r}}}$$
Let r be some multiple ##\eta## of ##r_s##, so that ##r = \eta \, r_s##.
Then we can write:
$$a = \frac{GM}{\eta^2 r_s^2\sqrt{1-\frac{1}{\eta}}}$$
and
$$\gamma = \frac{1}{\sqrt{1-\frac{1}{\eta}}}$$
Making use of the defintion of ##r_s##
$$r_s = \frac{2GM}{c^2}$$
and solving for ##\eta## as a function of ##\gamma##
$$\eta = \frac{\gamma^2}{\gamma^2 - 1}$$
we can re-write the expression for a as:
$$a =\frac{GM}{\eta^2 r_s^2\sqrt{1-\frac{1}{\eta}}} = \frac{GM\gamma}{\eta^2 r_s^2} = \frac{GM}{r_s} \frac{\gamma}{{\eta^2 r_s}}=\frac{c^2}{2} \frac{\gamma}{\eta^2\,r_s} = \frac{c^2}{2\,r_s} \gamma \left( \frac{\gamma^2-1}{\gamma^2} \right)^2 $$
The term to focus on is ##c^2/2r_s##. For ##\gamma=2##, the term on the right side turns out to be 18/16, so it's close to 1. For higher values of ##\gamma##, the term becomes approximately equal to ##\gamma##.
A large r_s makes ##c^2 / 2\,r_s## smaller, to minimze crush we want it as small ass possible. So we want a large black hole. Suppose we have a billion solar mass black hole. This is larger than the one at the center of our galaxy, which is about 4 million solar masses. The Scharzschild radius for one solar mass is about 1.5 km
<<wiki link>>, so for our billion solar mass black hole, we'd have ##r_s## = 1.5 billion km = ##1.5\,10^{12}## meters.
Plugging this in, we get ##\frac{c^2}{3\,10^{12} meters}## = 30,000 meters/second^2. Ouch. Assuming I didn't make any mistakes, something I'm not as good as avoiding as I used to be :(.This is why I say it's time for plan B when trying to get large time dilation factors. Hovering squishes you flat, trying to orbit the black hole runs into two problems. The first problem is that orbits don't exist inside the photon sphere, which is r=3m. One can get high time dilation due to velocity as one approaches the photon sphere, but one can get the same time dilation with high velocity outside the photon sphere. Furthermore, around a Schwarzschild black hole, the orbits at the photon sphere are unstable, so you'd need active correction to go this route. Not only does the gravity of the black hole not help much to achieve time dilation, it makes surviving the experience more difficult.
An interesting and non-obvious take of this is from Kip Thorne, the science consultant for "Interstellar". Tasked with finding a stable orbit with a high time dilation, he envisioned a very rapdily rotating black hole at the center, an came up with a possible (though not very likely) scenario that had some basis in physics that would work for the movie. But this isn't a Schwarzschild black hole at all, it's an extremal Kerr black hole. I rather suspect that in Thorne's scenario, the gravity also turns out to be much less important than the velocity, but I haven't actually computed anything.