When being reflected by a mirror, the photons do not lose "a tiny bit" of energy. Either they are reflected unchanged, or they are completely absorbed. A good mirror will reflect most of the photons, but will absorb a small fraction of them as well, say $0.1\%$ of them.
That is: Your photons don't lose energy over time; what happens is that the room loses photons over time: For each time a photon hits a wall, there is some probability $p$ that it will get absorbed ("consumed"). The chance that it doesn't get consumed after $N$ hits is $(1-p)^N$. Since the photons are very fast, they'll bounce off the walls very often in a short amount of time, so $N$ becomes really large really quick, and then $(1-p)^N$ becomes really small pretty fast, so after a short amount of time, all photons have been consumed with very high probability.
An important property of photons, that might not be entirely intuitive when coming from a "wave" background: The energy of an individual photon is determined entirely by the frequency of the light. Blue photons have higher energy than red photons. The intensity of light is determined not by the energy of your photons, but by their number.
If the photons would lose energy each time they bounce off a mirror, the reflections would change their color gradually, so that eventually blue light becomes red, then infrared etc. That doesn't happen: The mirrors don't change the color of the light. They only swallow some of the photons, i.e. they reduce the intensity.
With perfect mirrors, you could indeed expect to never lose any photons. But since anyone looking at the photons also absorbs then, the room would still get dark eventually. Unless there's nobody in there.
To be overly pedantic: Unless you keep the walls at zero temperature, you will always have some photons in the room as blackbody radiation. At "normal" temperatures, these photons are mostly in the infrared range, but if you make it really hot, the walls will start glowing.
Regarding the question "how to get light into the resonator in the first place":
It is done through the mirrors of the resonator. All mirrors are transparent to a certain degree, e.g. have high reflectivity but also a certain amount of transmittivity (e.g 0.1%).
The amount of bounces is usually only an effective one.
To understand how the number of bounces is measured consider the following:
Assume a narrowband light source (e.g. a laser) that shines light onto the back of one of the mirrors. Assume also that the wavelength of this light matches a cavity resonance (that is the cavity length of a multiple of the half wavelength). Assume further that one waits long enough that a steady state is achieved. Since the second cavity mirror is transmittive to certain degree as well, we see also light "leaking" out of the second mirror.
If one now immediately switches off the light source, then one sees a slow decay from this second output (high Q cavities achieve 100 micro-seconds maximum). We dont see an instantaneous turn-off at the output, because the light energy that is stored in the resonator is released "slowly" through the mirrors.
This decay time (or ring-down time) is a measure for the effective number of bounces.
Best Answer
The answer depends on how the periscope is made (the relative size of the mirrors, the shape of the box , and where do you put your eye. You do not need a periscope to see this, just look through a pipe. The mirrors do not matter they only redirect the light, so your problem is equivalent to that of making holes of the size of the mirrors at both ends of a box, instead of at the sides. Or you can replace the box by a pipe, to make it simpler. Basically, you will notice that depending on where you put your eye, you will see the inside or the outside of the pipe plus the hole at the other end.