Hydrostatic pressure on a teapot spout The phenomenon where water flows on the outside side of a teapot spout is named "The teapot effect", and occurs due to a difference in pressure between water and the atmosphere. Consider the image of a teapot spout below, and consider the pressures at points A, B, C and D.

(Quick translation: "Bico" means "Spout", and "Água" means "Water")
I've been kind of lost in this question due to different theories, so I'm going to arrange them into statements to make it more organized:


*

*(This doesn't alter the answer, but I wanted to know if it's correct.) The whole stream of water has a lower pressure (if compared to atmospheric) because it has a certain velocity, and fluids with velocity have lower pressure;

*A and D must be atmospheric pressure, since they're both in direct contact with air;

*B must be higher than both  because of Pascal's Law;

*C must be lower than D, but I don't have a clue why this is true. I'm assuming that because it must be the pressure difference that supports water against gravity (considering that I'm right about D being atm pressure), but that is kind of "doing the problem in reverse", what would the real reason be?
 A: This answer is a bit of a long story, but I have split it up for the different statements for your convenience. Having thought about it a bit more after the discussion with @Mephisto I actually believe that Bernoulli's equation is not applicable in points B and C, because it is based on conservation of energy and therefore only applies if wall friction is negligible.
1: False
The pressure at point D will be higher than atmospheric (see answer for statement 2). Given that the liquid is connected to the same reservoir (the tea inside the teapot), Bernoulli's equation would predict that in the low velocity regions, i.e. points B and C, the pressure will be higher according to:
$$\frac{1}{2}\rho v^2 +\rho g h + p = \text{constant}$$
However, it should be noted that Bernoulli's equation is not valid in points B and C, because we are in the laminar boundary in which frictional losses are non-negligible. Given that losses to the wall will lower the pressure we don't know whether the pressure is above or below atmospheric in these points so the only statement we can make is:
The pressure in the liquid is at best only at SOME places lower than atmospheric, but not at ALL places
2: False
For the interface of two immiscible fluids (gas-liquid or liquid-liquid) there will be a pressure jump across the interface depending on the curvature of the interface. This is predicted by the Young-Laplace equation:
$$\Delta P_c =\gamma \left(\frac{1}{R_1}+\frac{1}{R_2}\right)  $$
where $\Delta P_c$ is the capillary pressure jump across the interface, $\gamma$ is the surface tension between the two fluids and $R_1$ and $R_2$ are the two principle radii of curvature. In case of the water running along the bottom of the spout, say at point D, we have by approximation the following situation:

In this case the first radius of curvature $R_1$ has some value whereas the second radius of curvature $R_2$ goes to infinity (straight lines have no curvature). The pressure on the convex side (so in the liquid) will be higher than the pressure at the concave side (gas side) thus the pressure in point D is above atmospheric. This is in fact necessary to keep the liquid from falling down.
For point A I am actually not completely sure. In the drawing of the OP the interface also looks convex there, but I think this is more of an artist impression than the truth. If the interface is flat it would mean that the pressure at point A is atmospheric.
3: True
The pressure in B should indeed be higher than in point A and D because of Pascal's law and the simple fact that the flow is running from A/B towards D. This last addition is necessary, because with sufficient pressure difference it is obviously possible for a liquid to flow up against gravity (e.g. by pumping).
4: True
Indeed the pressure in C should be lower than in D (or at least equal to) otherwise the liquid would simply flow from C to D and thus detach from the solid. Here, the explanation is the adhesion of the solid and the liquid as explained in great detail in this paper (later published in Phys. Rev. Lett.).
In short, if the solid is hydrophilic ('likes' water) there will be an attractive force pulling the water against the solid thus introducing a lower pressure near the wall than further away from it. When the surface becomes more and more hydrophobic, the effect slowly disappears until, at the extreme of full hydrophobicity, it completely disappears according to:
$$We_{crit} \propto \left(\frac{r_i^2}{e_0^2}+\frac{r_i}{2e_0} \right)\left(1+\cos \theta_0\right) $$
where $We_{crit}$ is the critical Weber number (which thus relates the velocity as well), $r_i$ is the radius of curvature at the spout, $e_0$ is the thickness of the liquid stream and $\theta_0$ is the contact angle which indicates the wettability ($\theta_0=0$ for complete hydrophilicity and $\theta_0=180$ for complete hydrophobicity).
A: Warning: Another user has given a better answer. This one was chosen as the best one, before the other answer was written.
First, a simplified approach based on Bernouilli's equation for incompressible fluids:
Points B and C are directly in contact with the surface of the tube, thus they are nearly at zero speed with respect to it. But the fluid in A and D moves faster. Bernouilli's equation between the (nearly) quiet surface of the tea inside the teapot, and point A (or D) that is moving fast, tells you that the pressure in A (or D) is lower.
That is not the case with B (or C) since, as said, they are at zero speed with respect to the spout surface.
Bernouilli's equation is equivalent to a conservation of mechanical energy, for a small piece of fluid that translates from one point at pressure $P_1$ and speed $v_1$ to a second point with different values:
$$\tfrac12\, \rho\, v^2\, +\, \rho\, g\, z\, +\, p\, =\, \text{constant}$$
This equation holds along a streamline; in practical terms you can apply it between two points at a certain distance along the flow. One of them, as said, at the surface of the tea inside the teapot ($v=0$, $P= 1atm$) and the other may be A ($v>0$, therefore $p<0$).
Statements 1, 3 and 4 seem correct, although they can be fine-tuned: 1 and 4 agree with Bernouilli's equation, and 3 is true, but not only Pascal principle applies here, but also Bernouilli's equation tells you that the slower fluid at B must have a higher pressure.
As you see, the fact that the fluid is stopped at the layer directly in contact with the surface of the tube is very important, and so this effect is more likely to happen with some materials. You surely have noticed too, that when you give the fluid in general a high speed (by bending the teapot), it enters the turbulent regime and that zero speed layer (together with the embarrassing problem of dropping the tea out of the cup) disappears.
This link is interesting as well.
Regarding point 2, there I had a short but interesting chat discussion with @michielm, from which I learnt something new regarding what happens in the surface between air and tea. You should look at the links for more information.

PS: The OP and I crossed a couple of comments about the singer in his avatar picture, Carré Callaway, but they have disappeared. Do comments get deleted automatically or something?
