or some very big container with water, so big that its volume is not
changing
Yes, let's do that: imagine the left-hand side of the pipe to be connected with an infinitely large reservoir.
The fluid is considered both inviscid and incompressible.
Let's also make the height difference $h_1$ between the surface of the reservoir and the centre line of the (horizontal) pipe, such that with Pascal's Law we have:
$$p_1=p_A+\rho g h_1$$
Now we imagine a flow line that connects a point on the surface of the reservoir with a point right on the centre of the pipe and at its entrance. We can apply Bernoulli and because the flow speed $v_1$ at the surface of the reservoir is $\text{zero}$, so that with Torricelli's Law (basically a simple application of Bernoulli's principle) for inviscid fluids we get for the entrance speed $v_{ent}$:
$$v_{ent}=\sqrt{2gh_1}$$
Because the fluid is incompressible and the pipe's cross-section is constant, then per the continuum equation the flow speed is constant along the flowlines of the entire length of pipe. I'll call that velocity $v_2(=v_{ent})$.
Now imagine a flow line going from the centre of the entrance of the pipe to the top of the manometer tube (where you write $h_2$ and $p_2$). $p_2$ (top of the manometer tube) is of course $p_A$ and the flow velocity is $\text{zero}$ at that point.
We apply Bernoulli between these points:
$$p_1+\frac12 \rho v_{2}^2=p_A+\rho g h_2$$
$$p_A+\rho g h_1+\frac12 \rho v_{2}^2=p_A+\rho g h_2$$
$$\rho g h_1+\frac12 \rho v_{2}^2=\rho g h_2$$
So that:
$$\boxed{h_2=h_1+\frac{1}{2g}v_{2}^2}$$
Due to the nature and spirit of the question I've answered it using 'typical' approximations, used to approach such problems.
If one inserts the value of $v_2$ into the last equation, one obtains $h_2=2h_1$, which doesn't sound realistic.
There are two approximations in particular which deserve some scrutiny and both derive from the inviscidity assumption.
Firstly, when a viscous fluid enters a pipe the way it does in this problem, then the sudden contraction of flow causes pressure loss ('head loss', for engineers). This reduces $v_2$ somewhat.
Secondly, flow of a viscous fluid through a pipe, whether laminar or turbulent also causes pressure loss. So by the time flow reaches the point $2$, there is further loss of pressure.
If the sum of both pressure losses is written as $-\Delta p$, then the equation for $h_2$ could be rewritten as:
$$h_2=2h_1-\frac{\Delta p}{\rho g}$$
But it would be impossible to estimate this pressure loss without additional data such as fluid viscosity, pipe diameter, pipe length, position of point $2$ and pipe relative roughness.
