As it happens, I just recently went through that calculation myself for a different site.
Given the following facts from a quick web search, it isn't difficult to work out the numbers.
- The maximum efficiency of a (large) windmill is about 40%.
- The density of air is 1.225 kg/m3
- You need about 50 mW (10 mA at 5V) to light up an LED
First, we'll need about 50 mW / 0.40 = 125 mW of air power flowing through the windmill to create the electricity we need (ignoring other factors such as the actual efficiency of a small windmill and the efficiency of the generator).
The power of the air flowing through the windmill is 0.5mv2, where m is the mass rate of the air flowing through the "disk" defined by the diameter of the blades. For example, suppose we have a disk of 0.03m2 (about 20cm in diameter). The mass rate of the air is the area of the disk multiplied by the air velocity, multiplied by the density of the air:
$$\text{Mass rate} = 0.03 \text {m}^2 \cdot v \cdot 1.225 \text{ kg/m}^3 = v \cdot 0.03675 \text{ kg/m}$$
The power of that air is therefore:
$$P = 0.5 \cdot \text{Mass rate} \cdot v^2$$
Substituting and solving for $v$:
$$v = \sqrt[3]{\frac{0.125 \text{ W}}{0.5 \cdot 0.03675 \text{ kg/m}}} = 1.9 \text{ m/s}$$
... or about 7 km/h.
Taking into account the efficiencies we ignored earlier, plus the losses in a gear train that might be needed to get the generator RPMs up to a usable level, I would probably shoot for about 4× the area, or about 2× the diameter (40-50 cm), in order to get reasonable results.