Does the weight of an hourglass change when sands are falling inside?
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.
Imagine an hourglass with just one stone inside. When the stone start to falling a scale would stop to measure it's weight, but it will measure a spike corresponding to the moment when it hits the bottom. The bigger the airtime, the bigger the spike. It is like concentrating the weight of the stone in a very specific time interval: when it hits. However the average weight over the falling time is exactly the total weight with the stone at the bottom.
Going back to sand the only thing that change is that instead of a big spike, you have a lot of small spikes in the weight. So you don't have to wait a falling time to get the static weight as an average and a scale will average by itself inertia showing always the same weight. However if you find a scale with an outstanding resolution both in mass and time according to the size of the grains, you may be able to see these spikes.
Well, how to prove that we are just concentrating the weight in time. I think a quite simple, still effective argument is found in the high-school-level relation:
$$m \Delta v = F \Delta t$$
The momentum acquired by the rock during the free falling is:
$$M_f = m g T$$
during the impact the velocity is killed in a time $t$ and so the momentum, calling $a$ the involved acceleration we have (modulo any signs which are trivial to fix):
$$ m g T = m a t $$
It's now clear that
$$ m a = m g \frac{T}{t} $$
This means that, if for a time $T$ we are not measuring the stone weight, then for a time $t$ we measure a weight $\frac{T}{t}$ times bigger. The average in time equals to:
$$\frac{0\cdot T+mg\frac{T}{t} \cdot t+mg\cdot t}{T+t} = mg$$
The first term is related to the time of flight (zero force), the second is the force that kills the momentum over a time $t$ and for the same time $t$ also the natural weight of the rock acts.
If there is some air resistance, it will transmit some force to the scale while the stone's falling, in the average expression it will move some force from the second term to the first. The idea is that during the falling time the scale measures the drag, but then the velocity is a bit smaller when the stone hits. This would be proven in a similar way as above.
Actually sources of weight change has to be searched in the famous equation: $E=mc^2$. The energy of the sand down, will be a little less and so it's mass.
At the same time one could consider that going closer to the Earth surface the gravity is a little bigger, so sand down weight a little more.
Both the effects are far from measurable.
The apparent weight is indeed larger when the hourglass is running than at rest. See here for a detailed write-up. This effect has even been verified experimentally.
In a nutshell: the net effect of the flow is to move sand from the top surface (where it has a downwards velocity $v$) to the bottom pile, at rest. Thus, the sand is decelerating and the force on the balance is higher than at rest.