Wow, 700 feet, that's over an eighth of a mile! Because of the cost to double the diameter of the wire becomes significant at this length, consider a higher voltage mill if possible. The power consumed in getting the power to your house is equal to the square of the current multiplied by the resistance in the wire(I^2*R). Since the resistance is generally a function of the square of the diameter, doubling the diameter of the conductor increases the area by 4 times and cuts the resistance to 1/4th of its original value. Doubling the voltage for the same amount of power cuts the current in half, this results in the power loss (I^2*R)being cut to a fourth of its original value as well. If you havn't already gathered the components for your system, starting with a 48V system is probably the cheaper route. The longer the wire runs, the more advantage to a higher voltage system.
An example comparing doubling the system voltage or the conductor size if you had 24 watts of power from a fixed power source. (a little different than a wind turbine, but suitable for an example)
Current = I
Source Voltage = V
Conductor Resistance = R
Power Consumed in conductor = P (this is what we want to minimize)
base example: I = 1 Amp, V = 24 Volts, R = 1 Ohm, then P=I^2*R = 1 watt
effect of doubling conductor diameter with everything else the same as in base example: I = 1 Amp, V = 24 Volts, R = 1/4 Ohm, P = I^2*R = 1/4 watt
effect of doubling source voltage with same power from source and other factors same as in base example: I = .5 Amps, V = 48 Volts, R = 1 Ohm, P=I^2*R = 1/4 watt
You could locate the inverter and batteries near the tower and transmit 120 or 240 volt AC to the house. This may make system maintenance/monitoring more difficult, and a minor consideration is that the byproduct of rectifying/transforming the electrical energy is heat, although probably only a few bucks worth of equivelant electric heat over a heating season and not much of an issue at all if you heat with wood. Rich Hagen