There are too many unknowns to model the situation with any accuracy.
If you know the power being pumped into your pan then you can easily calculate the amount of steam generated from the latent heat of vaporisation of water. You can calculate the power being generated by your cooker from the flow rate of the gas, or the current if it's an electric cooker, but it's anyone's guess what percentage of this is lost to the environment and how much ends up in the pan.
For any significant flow rate of steam I would guess that turbulent mixing will ensure the steam temperature is roughly constant throughout the pan. The temperature will be whatever the boiling point of water is at the internal pressure. The steam flow rate through the hole in the pan is fairly straightforward to calculate from first principles, though since steam is so important industrially I'd guess some Googling will find tables and empirical equations for flow rate.
Response to comment:
Let the mass of water lost per second be $m$, then the power applied to the water in the pan is just:
$$ W = mL $$
where $L$ is the latent heat of vaporisation of water. This will be equal to the power generated by your cooker times some unknown factor less than unity to allow for heat loss to the environment.
Response to second comment:
The temperature of the water will be close to the boiling point because any water hotter than the boiling point turns to steam, and the latent heat required will cool the water again. The steam in the layer immediately above the water will be at the same temperature as the water because it's in thermal contact with it.
If the steam above the water is hotter than the water you have to ask what is heating it. The only things I can think of that could heat the steam are the pan walls and lid. However the pan is only being heated from the bottom, and heat flow by conduction though the pan walls is a lot slower than heating/cooling by convection between the pan walls and the water in the pan. Therefore I would guess that the pan walls and lid are also close to the boiling point of water - actually they will probably be slightly cooler because they will lose heat to the surrounding air.
So I would guess that as long as there is enough water in the pan the steam in the pan will be close to the temperature of the water.
When two phases are in equilibrium the chemical potential of the atoms/molecules in the two phases are the same. If you're not familiar with the concept of chemical potential it basically just means that the molar Gibbs free energies of the two phases are equal so the $\Delta G$ for the phase change is zero.
The argument is:
If the liquid and vapour are in equilibrium then the chemical potential in the liquid $\mu_l$ and vapour $\mu_v$ must be the same: $\mu_l = \mu_v$.
If the solid is also in equilibrium with the vapour then the chemical potential of the solid $\mu_s$ and vapour $\mu_v$ are also the same: $\mu_s = \mu_v$.
And that means that the chemical potential of the liquid and solid must be the same, because both are the same as the vapour: $\mu_s = \mu_l = \mu_v$.
And finally if the chemical potential of the liquid and solid are the same then it means the liquid and solid are in equilibrium i.e. we are at the freezing/melting point.
Re the second paragraph: when the solvent freezes it will freeze to form pure solid solvent i.e. it excludes the solute. So we have pure solid solvent in equilibrium with the vapour and "solvent + solute" in equilbrium with the vapour. So it's the same argument as above.
Best Answer
You can have a look at this pressure/temperature phase diagram of water:
Phase diagram taken from Martin Chaplin's webpage, under license CC-BY-NC-ND. This webpage is highly recommended, with tons of useful links and articles.
For reference, the diagram shows a point labeled $`` \textbf{E} "$ for fairly standard human conditions, around $25\sideset{^{\circ}}{}{\mathrm{C}} ~ \left(\sim 77\sideset{^{\circ}}{}{\mathrm{F}}\right)$ and normal atmospheric pressure.
The phase boundary between $\color{darkblue}{\textbf{Solid Ih}}$ and $\color{green}{\textbf{Liquid}}$ represents the temperature/pressure combinations at which water coexists between liquid water and solid ice. This boundary shows that the freezing temperature is roughly constant over a large pressure range, from about the triple-point (where solid, liquid, and vapor can coexist) and up to a pressure of about $200 \, \mathrm{MPa} .$