That is exactly right. A fundamental tenet of physics is that all inertial reference frames are equivalent and indistinguishable.1 Furthermore, given one inertial frame (standing at rest2), any other frame moving with respect to it with a constant velocity is also inertial. The frame "moving at terminal velocity" is just as inertial as "sitting still" and so you would not even be able to tell you were moving.
By definition you feel no acceleration at constant velocity. Thus the acceleration due to gravity must be exactly balanced by some other force. By construction that force is not air resistance for you (as would be the case of a sky diver at terminal velocity) but simply the normal force of the elevator floor, which would make the experience feel exactly like standing in a non-moving elevator in the same gravitational field.
1 At least locally, meaning that any experimental apparatus and things you measure are confined to objects also in that frame.
2 To be pedantic, standing "still" in a gravitational field is considered inertial in Newtonian mechanics but not general relativity. I am speaking in Newtonian terms here, but the conclusion would be just the same if analyzed with the machinery of GR.
Analyzing the acceleration of the center of mass of the system might be the easiest way to go since we could avoid worrying about internal interactions.
Let's use Newton's second law: $\sum F=N-Mg=Ma_\text{cm}$, where $M$ is the total mass of the hourglass enclosure and sand, $N$ is what you read on the scale (normal force), and $a_\text{cm}$ is the center of mass acceleration. I have written the forces such that upward is positive
The center of mass of the enclosure+sand moves downward during process, but what matters is the acceleration. If the acceleration is upward, $N>Mg$. If it is downward, $N<Mg$. Zero acceleration means $N=Mg$. Thus, if we figure out the direction of the acceleration, we know how the scale reading compares to the gravitational force $Mg$.
The sand that is still in the top and already in the bottom, as well as the enclosure, undergoes no acceleration. Thus, the direction of $a_\text{cm}$ is the same as the direction of $a_\text{falling sand}$ . Let's just focus on a bit of sand as it begins to fall (initial) and then comes to rest at the bottom (final). $v_\text{i, falling}=v_\text{f, falling}=0$, so $a_\text{avg, falling}=0$. Thus, the (average) acceleration of the entire system is zero. The scale reads the weight of the system.
The paragraph above assumed the steady state condition that the OP sought. During this process, the center of mass apparently moves downward at constant velocity. But during the initial "flip" of the hour glass, as well as the final bit where the last grains are dropping, the acceleration must be non-zero to "start" and "stop" this center of mass movement.
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It depends on how hard you can push. Realistically, it would not have much effect. After all, the height at which you can throw a person is negligible compared to $10$ meters. Btw, at such heights, you can use the free fall equation and get the impact velocity of roughly $35$ km/h, so it is survivable as expected. My guess is that if the other person is trying to preserve you, his best strategy would be to hug you to cushion your fall. He can't change the impact velocity, but by lengthening the deceleration phase (which the most vital part) it might be worth it at his expense. Hope this helps.