Helium balloons are pulled by gravity, as are all objects with mass. The reason they don't fall is that there is another force acting on them, a buoyant force from air pressure that is equal to the weight of the air displaced by the balloon.
The reason you don't float is that the weight of the air you displace is quite a bit less than your weight (a person is more dense than air). The reason a normal balloon doesn't float is that the weight of the air it displaces is just a little bit less than the weight of the balloon (because it is filled with air, but the rubber of the balloon itself is more dense than the air).
The analogy you want is to objects floating (or suspended) in water. Most rocks sink to the bottom, pulled by gravity, because the weight of the water they displace is less than their own weight. A bowling ball (ironically) is very close to the same density as water, so it will float suspended in mid-water, just like the helium balloon that has leaked a little bit.
The upthrust on the balloon is equal to the weight of air displaced, so we get:
$$ F = V_b \rho g \tag{1} $$
where $V_b$ is the volume of the balloon and $\rho$ is the density of the air. Assuming air is approximately an ideal gas it obeys the equation of state:
$$ PV = nRT $$
so the molar density is:
$$ \rho_M = \frac{n}{V} = \frac{P}{RT} $$
where $n$ is the number of moles of air. The density in kg/m$^3$ is given by multiplying the molar density by the (average) molar mass of the air $M_{\text{air}}$, and substituting this in equation (1) we get:
$$ F = V_b M_{\text{air}} g \frac{P}{RT} \tag{2} $$
Now let's consider what happens to the volume of the balloon. We'll take the two extreme cases where the rubber skin is infinitely rigid and where it's infinitely compliant.
First consider the case where the rubber skin is infinitely compliant i.e. it doesn't exert any force on the helium inside it. In that case the volume of the helium is (approximately) given by the ideal gas equation:
$$ V_b = \frac{n_{\text{He}}RT}{P} $$
where $n_{\text{He}}$ is the number of moles of helium. Substituting this into equation (2) we get:
$$ F = n_{\text{He}} M_{\text{air}} g $$
which is constant. So in this case we find that the bouyancy is unaffected as the pressure and temperature change.
Now consider what happens if the rubber skin is infinitely rigid, in which case the volume $V_b$ is constant. We end up with:
$$ F \propto \frac{P}{T} $$
In this case the bouyancy is affected by the pressure and temperature. Assuming the pressure is approximately constant the bouyancy is inversely proportional to temperature so the balloon will rise when it gets cold and fall when it gets hot, which matches your observation.
I've taken the two extreme cases because I don't know the equation for the force produced by the rubber skin of the balloon. However it is presumably somewhere in between the two extremes I've discussed, so we expect the behaviour of the balloon to fall between those two extremes. That means we expect the bouyancy will increase as the temperature decreases and vice versa.
Best Answer
One indication whether the 'inner' balloons are filled with helium or something heavier is whether the 'outer' balloon still floats in air or not.
Whether it floats or not depends on whether its overall density is lower than air's density or not. Its overall density $d$ is simply given by:
$$d=\frac{\Sigma m}{V}$$
Where $\Sigma m$ is the sum of all the masses that make up the balloon and $V$ its total volume.
The sum of all masses means the sum of:
If the inner balloons are filled with air and there are too many of them the overall density will exceed that of air and the outer balloon will not float.
But even if the inner balloons are all filled with helium, the outer balloon may still not float. That's because a larger number of inner balloons making up the same volume of a smaller number of inner balloons with create more mass because of the increased surface area and the inherent 'cost' in mass of larger number of inner balloons. For that reason:
... is not necessarily correct.
If the mass of the outer balloon material, the total volume of the balloon, the mass of the inner balloon material and the volume of the inner balloons was known, then depending on whether the outer balloon still floats in air or not, the lower or lower limit of the density of the inner balloon gas could be calculated.
Call $m$, $d$ and $V$, with resp. suffixes $o$ and $i$ for outer and inner balloons, the mass of balloon material, the density of the filling gas and volume of the balloons. Let there be $n$ identical inner balloons. The overall density of the balloon is then:
$$d=\frac{m_o+d_o(V_o-nV_i)+nm_i+nd_iV_i}{V_o}$$
If the balloon just about floats in air, then:
$$d=d_{air}$$
From which we can deduce:
$$d_i=\frac{d_{air}V_o-m_o-d_o(V_o-nV_i)-nm_i}{nV_i}$$