You must be aware that when two containers of water are connected by a pipe and if those containers are not airtight, water will flow between them until water-levels become equal. Obviously the container for feeding birds is open to atmosphere.
Now if the reservoir is not airtight (perhaps there is a hole or cut somewhere, through which air can flow in directly from ambient to reservoir) then the necessity for equalization of water levels explains the results of experiments #2, #3, and #4. It can also explain results of experiment #1 if water in the reservoir wasn't initially deep enough to cause overflow from the container, when water-levels equalized.
But say you have been extremely careful to ensure that there is no hole or cut on the reservoir that can allow air in directly from ambient to reservoir. In this case what you have observed can still happen, and may be qualitatively explained thus: As water is emptied out of the airtight reservoir the air initially inside the reservoir (atop the water surface) expands resulting in a lower pressure there. Therefore the body of water inside the reservoir experiences two counter-forces: one is due to the hydrostatic head caused by the difference in water levels inside the reservoir and the container and which causes the water to flow out of the reservoir; and second is due to the difference in ambient pressure outside and low pressure zone inside the reservoir atop the water surface that prevents the water from flowing out. At some level difference between the water in the reservoir and the container (and for a given geometry of the reservoir), the two forces become equal and flow shall stop. However by the time this point of no-flow is reached the container might already be overflowing.
This idea may be formulated precisely. Let $V_0$ be the initial volume of air inside the reservoir atop the water surface, which is also at atmospheric pressure. Let us measure all heights from the ground on which the reservoir and the container rest. Let us take the container to be sufficiently deep that water cannot overflow. When water flows between them and equilibrium is reached, let $h_R$ and $h_C$ be the water levels in the reservoir and container respectively, measured with respect to the ground. Corresponding to final water level in the reservoir $h_R$ the volume of air atop the water surface in the reservoir shall be $V$, so that air in the reservoir has expanded from $V_0$ to $V$. If the temperature change is insignificant so that we may assume it to be constant, using ideal gas law for air (which is a good approximation) we have magnitude of pressure difference w.r.t. ambient as:
\begin{align}
|\Delta p|=mRT\left( \frac{1}{V_0}-\frac{1}{V} \right)
\end{align}
in which $m,R,T,$ are respectively mass, gas constant, and temperature of air inside the reservoir. In equilibrium this pressure difference is counterbalanced by the hydrostatic head $\rho_{water}g(h_R-h_C)$. This gives:
\begin{align}
|\Delta p|=\rho_{water}g(h_R-h_C) & =mRT\left( \frac{1}{V_0}-\frac{1}{V} \right)\\
\therefore\quad h_C & =h_R-\frac{mRT}{\rho_{water}g}\left( \frac{1}{V_0}-\frac{1}{V} \right)
\end{align}
If the actual height of the container is less than $h_C$ then the water will overflow. Remember that $V$ in the expression above is a function of $h_R$ and each configuration in your experiments shall require a separate calculation.
P.S. I am not sure why you do not wish to employ valves. Float valves are not very expensive and do not require power input to work. There is another (not very pleasing to me) alternative to maintaining a nearly constant water level without using any extraneous devices. Take the container with as small a floor area as is feasible, and choose a reservoir with as large a floor area as is feasible. Then for a given amount of water flowing from reservoir to container the level change in the reservoir shall be smaller the larger the ratio of area of reservoir to container. To be precise, let $A_R,A_C$ be the area of reservoir and container respectively. Let $\dot{q}''$ be average rate at which volume of water is consumed from the container per unit area per unit time, due to birds drinking from it and due to evaporation. Let us assume that the rate at which the container loses water is small enough (which seems a reasonable assumption to me) so that water levels in reservoir and container equalize practically instantaneously, so that at all times water levels in reservoir and container are equal. Then $\dot{q}''A_C\delta t$ is the volume of water lost from reservoir in small time duration $\delta t$. This causes a dip in water level $\delta h$ in reservoir and container. Mass balance then gives: $(A_R+A_C)\delta h=\dot{q}''A_C\delta t$, therefore the rate at which height of water in container changes is: $\dot{h}=\dot{q}''/(1+A_R/A_C)$.
Water in a fluid phase has density of $1.00~\text{g/cm}^3$, while in a solid form (ice) - $0.92~\text{g/cm}^3$. So while freezing water expands in volume (and thus drops in density), it becomes lighter. That's why we see pieces of ice floating in a river. Due to same water anomalous property, deep lake bottoms may never get completely frozen, because liquid water heat going-up may never reach lake surface and may dissipate in the lower layers of ice. Thanks to that, fish may enjoy living in a winter seasons.
Thus, if your pool is deep enough it may never get frozen fully, or at least will do it slowly if you fill it in one go. A Better tactic is - freezing the ice layer by layer. So that you'll have total control over how thick the ice is. In addition to that, the heat amount contained in a thin layer of fluid water will be lower compared to the case of a fully filled volume, so this heat will escape to surface faster. My advice - fill thin layer of water in the bottom, wait until it freezes to ice. Then fill another layer of water on top of it and repeat the process until whole pit will be complete ice.
Best Answer
There are two effects that both reduce the final water level:
When the water is pouring into the bottle and back out of it, it does not immediately turn around at the surface. Instead, the kinetic energy of the water causes it to flow quite deep into the bottle, then make a turn and flow back upwards.
When the incoming flow stops, the remaining water in the bottle still has that kinetic energy, and will continue flowing upwards and over the rim for a short time.
Depending on the faucet, the water flow usually has also entrapped air bubbles, which make it appear white rather than clear. Once the flow stops, the larger bubbles quickly rise to the surface and pop, further lowering the water level.
Just for fun, I took a slow motion video of filling a bottle (slowed 8x). With my faucet, it appears the contribution of air bubbles is quite large.