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.
Best Answer
While everyone agrees that jumping in a falling elevator doesn't help much, I think it is very instructive to do the calculation.
General Remarks
The general nature of the problem is the following: while jumping, the human injects muscle energy into the system. Of course, the human doesn't want to gain even more energy himself, instead he hopes to transfer most of it onto the elevator. Thanks to momentum conservation, his own velocity will be reduced.
I should clarify what is meant by momentum conservation. Denoting the momenta of the human and the elevator with $p_1=m_1 v_1$ and $p_2=m_2 v_2$ respectively, the equations of motion are
$$ \dot p_1 = -m_1 g + f_{12} $$ $$ \dot p_2 = -m_2 g + f_{21} $$
Here, $f_{21}$ is the force that the human exerts on the elevator. By Newton's third law, we have $f_{21} = -f_{12}$, so the total momentum $p=p_1+p_2$ obeys
$$ \frac{d}{dt} (p_1 + p_2) = -(m_1+m_2) g $$
Clearly, this is not a conserved quantity, but the point is that it only depends on the external gravity field, not on the interaction between human and elevator.
Change of Momentum
As a first approximation, we treat the jump as instantaneous. In other words, from one moment to the other, the momenta change by
$$ p_1 \to p_1 + \Delta p_1, \qquad p_2 \to p_2 + \Delta p_2 .$$
Thanks to momentum "conservation", we can write
$$ \Delta p := -\Delta p_1 = \Delta p_2 .$$
(Note that trying to find a force $f_{12}$ that models this instantaneous change will probably give you a headache.)
How much energy did this change of momentum inject into the system?
$$ \Delta E = \frac{(p_1-\Delta p)^2}{2m_1} + \frac{(p_2+\Delta p)^2}{2m_2} - \frac{p_1^2}{2m_1} - \frac{p_2^2}{2m_2} .$$ $$ = \Delta p(\frac{p_2}{m_2} - \frac{p_1}{m_1}) + (\Delta p)^2(\frac1{2m_1}+\frac1{2m_2}) .$$
Now we make use of the fact that before jumping, the velocity of the elevator and the human are equal, $p_1/m_1 = p_2/m_2$. Hence, only the quadratice term remains and we have
$$ (\Delta p)^2 = \frac2{\frac1{m_1}+\frac1{m_2}} \Delta E .$$
Note that the mass of the elevator is important, but since elevators are usually very heavy, $m_1 \ll m_2$, we can approximate this with
$$ (\Delta p)^2 = 2m_1 \Delta E .$$
Energy reduction
How much did we manage to reduce the kinetic energy of the human? After the jump, his/her kinetic energy is
$$ E' = \frac{(p_1-\Delta p)^2}{2m_1} = \frac{p_1^2}{2m_1} - 2\frac{\Delta p\cdot p_1}{2m_1} + \frac{(\Delta p)^2}{2m_1}.$$
Writing $E$ for the previous kinetic energy, we have
$$ E' = E - 2\sqrt{E \Delta E} + \Delta E = (\sqrt E - \sqrt{\Delta E})^2 $$
or
$$ \frac{E'}{E} = (1 - \sqrt{\Delta E / E})^2 .$$
It is very useful to estimate the energy $\Delta E$ generated by the human in terms of the maximum height that he can jump. For a human, that's roughly $h_1 = 1m$. Denoting the total height of the fall with $h$, we obtain
$$ \frac{E'}{E} = (1 - \sqrt{h_1/h})^2 .$$
Thus, if a human is athletic enough to jump $1m$ in normal circumstances, then he might hope to reduce the impact energy of a fall from $16m$ to a fraction of
$$ \frac{E'}{E} = (1 - \sqrt{1/16})^2 \approx 56 \% .$$
Not bad.
Then again, jumping while being weightless in a falling elevator is likely very difficult...