The net pressure on the liquid is just the atmospheric pressure.
Pressure in a fluid acts in every direction, but as the point is on the surface, $\text{P}_{water}=h\rho g=0$ as $h=0$. So only atmospheric pressure will be acting on point A.
The height of the liquid column doesn't affect the pressure on top. Pressure in a liquid is affected by the weight of the liquid above it.
EXPLANATION
Consider this diagram, :
The pressure on a cross section inside the fluid that height is given by the weight of fluid above that height, divided by horizontal cross sectional area of the fluid column.
We get the pressure by $$P=\frac{F}{A}=\frac{mg}{A}$$
$$=\frac{\rho Vg}{A}=\frac{\rho (A\times h)g}{A}$$
$$\boxed{P=h\rho g}$$
When we calculate the pressure at the surface of the fluid, we take the weight of fluid above the cross sectional area. But as there is no fluid above the surface, $\text{Weight}=0,~~ \therefore \text{P}=0$
Yes, generally speaking it makes sense to talk about macroscopic averages such as pressure also for gases as they are sufficiently dense in most common situations on earth ($Kn \ll 1$). In this case the pressure emerges from elastic collisions as well as additional repulsive forces.
Your statements
"in a gas however the atoms/molecules interact only via collisions, which are rare - the more so the more the gas is diluted."
and
However, in the bulk of the gas the molecules do not interact - apart from the collisions that are typically neglected (although they are responsible for the establishment of the thermodynamic equilibrium). If we were to place a continuous surface anywhere within the container, there would be pressure exerted on this surface by the molecules scattered from it. However, in the absence of such a surface the molecules continue to move freely.
are only partially true, gases on earth are actually denser than you would believe: Under norm conditions there are around $2.69 \times 10^{19}$ molecules in a cubic centimeter of air, which makes particle collision fairly common. The underlying continuum assumption (the assumption that suitable averages on a larger scale - macroscopic values such as pressure - can be found) only breaks down in extremely diluted flows such as in atmospheric entries. Under these circumstances the definition of macroscopic values such as pressure indeed does not make sense.
Kinetic theory of of ideal gases
In order to quantify under which conditions this happens (meaning that the continuum hypothesis loses its validity) one has to turn to kinetic theory of gases. There, starting from considerations based on elastic inter-particle collision (this is only a viable assumption for light mono-atomic gases, for larger molecules one would actually have to assume also far-field interactions caused by repulsive forces) in highly diluted gases - by employing statistical mechanics - one can find an evolution equation for a particle distribution, the Boltzmann equation.
From there you can go on and make estimates about the number of collision per time and phase space element, given by the so called Stoßzahl. For a gas in equilibrium one could find the number of collisions by assuming a Maxwell-Boltzmann equilibrium distribution and integrating out.
Finally one can try to find equations that emerge on a larger time scale by performing a so called Chapman-Enskog expansion, a perturbation analysis. There one finds that for the limits of small ratios of the mean free path $\lambda$ and the characteristic length scale of a given problem $L$, the Knudsen number
$$ Kn := \frac{\lambda}{L} \phantom{spacespace} \frac{\text{mean free path}}{\text{representative physical length scale}}$$
the Navier-Stokes equations which were initially derived for dense fluids on a macroscopic level. In this context one can find that the average fluctuations of the relative velocity (the velocity of the individual fluid particles relative to the macroscopic flow)
$$ p = \frac{m_P}{3} \int |\vec \xi - \vec u|^2 f d \vec \xi,$$
correspond to the macroscopic pressure.
As a threshold to continuum based flow generally $Kn \lesssim 0.01$ is taken. In this limit macroscopic averages like pressure and viscosity make sense, they are sufficiently smooth and vary only slowly in time. This is by no means a strict boundary but rather an order of magnitude for which the aforementioned second-order expansion in a perturbation series still makes sense. For larger Knudsen numbers common continuum assumptions like the no-slip condition will not be accurate but instead one will have a certain slip-velocity at a solid boundary and said average will become too noisy to actually have physical meaning on the scale of interest.
For an ideal gas one can correlate the Knudsen number to two other common characteristic numbers, the Reynolds number $Re := \frac{U \, L}{\nu}$ - a measure for turbulence - and the Mach number $Ma := \frac{U}{c_s}$ - a measure for compressibility
$$ Kn \propto \frac{Ma}{Re} $$
For non-ideal gases one actually has to include far field interactions and inner degrees of freedom which makes the evaluation of the integrals quite tricky and leads to the kinetic theory of non-ideal gases.
Best Answer
The location depends on why you are measuring the pressure. There will be a process reason for the pressure measurement, and that will determine the location of the pressure measuring device. Ex: it's possible to infer tank level from pressure. In that case, you probably want the pressure measurement at the bottom of the tank. On the other hand, if you are worried about something in the vapor space of the tank, you want the pressure measurement on the top of the tank.