More light reflected off of double pane windows? In glass windows, you get light passing through and light reflecting off the front and back surfaces. Would you get more light reflecting off of triple or double pane windows? Because you have double or triple the amount of surfaces for the light to reflect off of. 
 A: You hit on half of the idea of optical thin film coatings. The other half is that light is a wave. Multiple reflections can interfere constructively or destructively. This can be used to make either an antireflection coating, or a mirror, or something more complex. 
This shows the idea of an antireflection coating. Attribution below. The coating is 1/4 wavelength thick. Light that bounces off the back surface travels an extra 1/2 wavelength. It emerges out of phase with the front reflection. The two reflections cancel. 

It doesn't really work on something as thick as a pane of glass. You can't make the front and back surfaces of glass exactly parallel and exactly the right separation. 
Also, as WetSavannaAnimal pointed out, ordinary light does not have a perfectly stable wavelength. It is produced by a lot of individual short lived events in many different atoms, and each atom is independent. By the time a beam of light travels through a pane of glass twice, the light arriving at the top surface is from different events. There is no consistent relationship in the phases of light separated that far. So you don't get a cancellation that lasts longer than a fraction of a nanosecond. 
A good quality laser has a very stable wavelength. With it, you can get interference patterns from the front and back surfaces of a pane. This is the original of the speckle that a laser produces when you shine it on a rough surface. You get interference from reflections from low and high points. 
Illustration by DrBob - Own work (enwiki), CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=12305780
A: Double pane reflects more light, because the second pane adds reflections and  light. 
It is best seen by using a point source, as in the photo. Image A0 is due to a single reflection in the first pane, B0 is due to a single reflection in the second pane, C0 is due to two reflections in the second pane. The other images are due to multiple reflections inside a pane. So the A-images are light from the first pane, and the B- and C-images would disappear if the second pane was removed. 

A: Short answer - yes.
If you have a single-pane window, light that doesn't reflect off the front or rear surface will be transmitted. But a second pane might once again reflect some of this light, and that has a chance of making it all the way out to the front. The second pane doesn't "stop" any of the reflection off the first pane - so it can only add to the total reflection.
Here is a diagram that demonstrates this:

Reflections A and B will happen with just a single pane; C and D are added with a second pane, and you can keep going.
Calculating the theoretical total reflection is not too hard. Assuming normal incidence, the Fresnel reflection equations tell us the reflected intensity is
$$r = \left|\frac{n_1-n_2}{n_1+n_2}\right|^2$$
From the symmetry of this equation you can see that the reflection off the back surface is the same as the reflection off the front surface - but then that ray (marked B) needs to be transmitted - where of course $t = 1-r$. 
This means that the intensity of A+B is given by $r+t^2r$ (A is just reflected, while B undergoes transmission twice and reflection once). For C, there are 4 transmissions and one reflection; for D, it is 6, etc. 
Thus, for a stack of N window panes, the reflected intensity (if there is no absorption, and light is too incoherent for interference effects to be significant), the reflected intensity is
$$I = r \sum_{i=0}^{2N-1} t^{2i}$$
This is a geometrical series that sums to
$$I = \frac{r~(1-t^{4N})}{1-t^2}$$
It is clear from this equation that as $N$ gets bigger, the reflected intensity also gets bigger - although it's a law of diminishing returns. We can plug in some numbers...
For a typical case of glass with n=1.5, we get $r=0.04$ and $t=0.96$, so the fraction of reflected intensity for a single pane (A+B) would be $0.04\cdot(1+0.96^2) = 0.077$; a second pane would bring the fraction up to 0.142, and an infinite stack would give a reflected fraction of 0.510
Peripherally related is this question and associated answer.
