What is the maximum height for a puddle of water, assuming STP? I wonder if anyone has figured this out. Assuming, standard temperature and pressure (273 K and 1atm), what is the maximum height of a water puddle on a flat surface? There might be some other factors I don't know about.
 A: The height of the puddle
I will use the common definition of puddle in the field of capillarity (which I believe you refer to) which is: a droplet on a flat horizontal surface flattened substantially by gravity as shown in the schematic below, coming from the book by De Gennes (2003). 
The droplet on the left is just a droplet (with contact angle $\theta_e$), the droplet on the right would be termed a puddle with the same contact angle (i.e. the same liquid on the same surface), but bigger in volume such that it is spread out to a size much larger than the capillary length $\kappa^{-1}$, causing it to be flattened by gravity to a maximum heigth $e$, which is the height you are asking for.
On the basis of a balance between surface forces and hydrostatic force (see below) it can be derived that this height $e$ depends on the capillary length and the contact angle as follows:
$$e=2 \kappa^{-1} \sin \frac{\theta_e}{2} \tag{1} $$ 
where $\kappa^{-1}=\sqrt{\gamma/\rho g}$
What you can see from this equation is that the height of the puddle depends on 3 (easily changeable) parameters: the liquid density $\rho$, the liquid-gas surface tension $\gamma$ and the contact angle $\theta_e$.
Conclusion
Given that $\gamma$ and $\rho$ and to a lesser extend $\theta_e$ as well, are dependent on temperature. You will need the appropriate values for STP conditions for water. Additionally, $\theta_e$ is dependent on the properties of the solid surface, so there you cannot tell the height of the puddle without knowing the surface it is laying on. And of course strictly speaking you need to know whether the droplet is on earth or some other planet.
The derivation
You can set up a simple balance of forces (per unit length) over a part of the droplet as shown below (again image from De Gennes (2003)).

So you have 3 surface tensions, $\gamma$, $\gamma_{SO}$ and $\gamma_{SL}$ for the three interfaces acting as shown by the arrows. Additionally you have a force $\tilde{P}$ from the hydrostatic head: $\frac{1}{2}\rho g e^2$.
Balancing those we get
$$\frac{1}{2}\rho g e^2+\gamma_{SO}-\gamma-\gamma_{SL}=0 $$
Plugging in Young's equation for the equilibrium contact angle: $\gamma \cos \theta_e=\gamma_{SO}-\gamma_{SL}$ we find
$$\gamma (1-\cos\theta_e)= \frac{1}{2}\rho g e^2$$
which upon rearranging becomes $(1)$
