There seems to be two questions here, how to evaluate thermoelectrics and how much cooling is required to freeze water. I'll lead off with some instruction on heat transfer and thermoelectrics and then discuss your design question.
Heat Transfer, Really Quickly
When heat conducts through a solid, the heat flow is proportional to the temperature difference such that $q = K \Delta T$, where $q$ is the heat rate (usually in Watts, energy moved per unit time); $K$ is the conductance, a property of the solid; and $\Delta T$ is the temperature difference
- Heat sink manufacturers will specify an effective conductance of their heat sinks, usually in °C/W
- The "R-value" of insulation is the inverse of conductance in really weird units
Thermoelectrics
Thermoelectric datasheets contain values that are easy to measure (for the manufacturers) but not easy to put into system models (for you). The first step is to get the Seebeck coefficient, $\alpha$, and the thermal conductance, $K$, from the data provided. These properties are important because they feed into the standard model of thermoelectrics1:
$$ q_{c} = \alpha I T_c - \frac{1}{2} I^2 R - K (T_h - T_c) $$
$$ q_{h} = \alpha I T_h + \frac{1}{2} I^2 R - K (T_h - T_c) $$
$$ V = \alpha (T_h - T_c) + IR $$
Where $q$ is heat rate (more on that in the design section), $I$ is current, $R$ is electrical resistance, $T_h$ and $T_c$ are the cold and hot side temperatures, respectively and $V$ is the applied voltage.
$K$ can be found easily by dividing Qcmax by Tmax (which properly should be called $\Delta T_{max}$). Tmax is the maximum temperature difference the module can sustain when the hot side is held at a reference temperature. At this point, all of the electrical power input is being used to pump the heat which was conducting through the device, i.e. $q_{c,max} = K \Delta T_{max}$
The seebeck coefficient can be found by using Qcmax. At max heat pumping, the temperature difference has to be zero (from the first equation), so $\alpha I_{max} T_c = q_{c,max} + \frac{1}{2} I_{max}^2 R$$
Designing a System
Your system, I imagine, will look a bit like this. Your thermoelectric will need to be able to pump out both the heat that comes in from the outside (how much will depend on how well you insulate your compartment) and whatever heat is released by the water during the freezing process.
As you noted, it takes 88 J/g of water to freeze water. If you multiply 88J/g of water by 10g (at 1g/mL) and divide by the number of seconds you want it to take, you will have your heat rate requirement $q$ to freeze the water. As Olin said, add in some amount of heat that will conduct through the walls and you have an estimate of $q_c$, the amount of heat the thermoelectric needs to move.
Finally, the heat sink that you buy will have some conductance rating $C$ in °C/W, we know that the hot side of the thermoelectric $T_h$ will be equal to $C q_h + T_{ambient}$.
Now we have enough information to solve the problem. Effectively, what we need to ask is whether there is a current $I$ that will satisfy the three equations below for the known $q_c$, $T_c$, $\alpha$, $R$, and $K$.
$$ q_{c} = \alpha I T_c - \frac{1}{2} I^2 R - K (T_h - T_c) $$
$$ q_{h} = \alpha I T_h + \frac{1}{2} I^2 R - K (T_h - T_c) $$
$$ T_h = Cq_h + T_{ambient} $$
Although you might be able to solve those equations by hand, I would just plug them into Excel and use the solver functionality to see if it is possible. If it isn't possible, try adjusting your $q_c$ requirement, e.g. by freezing the ice slower or adding a second thermoelectric module (which would cut $q_c$ per module in half.
