References for the terminal falling velocity of a skydiver give numbers in the range of $54 m/s$ to $76 m/s$. I expect that the real range is even larger, since it's strongly affected by the orientation and body position of the skydiver.
For normal atmospheric drag on this scale, we have the fairly good approximation that force is proportional to velocity squared, $F_{air} \propto v^2$. For normal terminal falling conditions, the upward air force is exactly equal to gravity, and the air force is a function of velocity exclusively because we're assuming the same air composition for now.
You ask about a fall that is "reliably survivable", I should note that this would still NOT be a comfortable fall. In low gravity environments, you could expect some gnarly bounces to top it all off. I don't have a good reference for this, but I would anecdotally put it in the neighborhood of a $40 ft$ fall, which would equate to about $15 m/s$.
I will refer to the air force during a normal terminal velocity fall on Earth as $F_{air}$ and introduce $F_{air}'$ for the force on the new planet. I'll do some simple equivalencies to get an answer for the new gravity needed.
$$ m g = F_{air} = (\text{const}) v^2$$
$$\frac{F_{air}}{F_{air}'} = \frac{v^2}{v'^2} = \frac{g}{g'}$$
$$\frac{v}{v'} = \frac{60 m/s}{15 m/s} $$
$$ \frac{g}{g'} = \left(\frac{60}{15}\right)^2 = 16$$
So my answer is quite simply that gravity would have to be 1/16th as strong as it is on Earth, or $0.6 m/s$. Are there any bodies in the solar system quite like this? Wikipedia is helpful here. Several bodies come close, like Pluto, Eris, or Triton, but none of them have much of an atmosphere. It is fun to think about, but I doubt that an atmosphere of such a high density with such a low gravity will be found in our local celestial neighborhood.
Walking would be difficult if such a planet existed, but not impossible. The moon is 1/3rd the gravity of Earth, so this hypothetical planet would be roughly 5 times less gravity than the moon. It would be very bouncy, but still very different from zero-gravity.
Terminal velocity is reached when the drag force due to moving through air is equal (but opposite) to the gravitational force. Now, the gravitational force is proportional to the mass, while the drag force has nothing to do with mass, but everything to do with how large and "streamlined" the object is. Suppose object A is twice as heavy as object B. If object A also experiences twice the drag force as object B (at a given speed), then their terminal velocities will be the same.
To put it another way, let's suppose that the two objects have the same masses, and therefore the same weights; they have the same gravitational forces. The question becomes: do they have the same drag force?
Drag comes from the resistance of the air to an object's movement, so – all else being equal – something that's more streamlined will have less resistance. If one of this is shaped like a bullet, and one is shaped like a big hollow ball, the big ball will have the same amount of drag at low speeds as the bullet does at high speeds. So the ball's terminal velocity will be a lot lower.
(I've simplified a little; the force of buoyancy should be added to the drag force. But usually this is relatively small, so we can ignore it just for simplicity.)
Best Answer
Terminal velocity is the (asymptotic) maximum velocity that you can reach during free-fall. If you imagine yourself falling in gravity, and ignore air resistance, you would fall with acceleration $g$, and your velocity would grow unbounded (well, until special relativity takes over). This effect is independent of your mass, since
$F = ma = mg \Rightarrow a = g$
Where terminal velocity arises is that air resistance is a velocity-dependent force acting against your free fall. If we had, for example, a drag force of $F_D=KAv^2$ ($K$ is just a constant to make all the units work out and depends on the properties of the fluid you're falling through, and $A$ is your cross-sectional area perpendicular to the direction of motion) then the terminal velocity is the velocity at which the forces cancel (i.e., no more acceleration, so the velocity becomes constant):
$F = 0 = mg - KAv_t^2 \Rightarrow v_t=\sqrt{mg/KA}$
So we see that a more massive object can in fact have a larger terminal velocity.