Get rid of the output capacitor. That circuit was probably meant to produce a signal around zero, so the capacitor is there to block the 1/2 Vdd offset. However, the microcontroller wants to see the signal centered around 1/2 Vdd, so just get rid of the capacitor.
Microcphones do need a lot of gain. Electrets can be sensitive, but you still might need a voltage gain of 1000. The gain in your circuit is the ratio of R5 to R2, but this only works within limits of what the opamp can do.
The values you mentioned above would give you a gain of 5000. That's a lot more than you should try to get from a single opamp stage. Not only will the offset voltage be multiplied by this gain, but the opamp won't be able to provide that over the full frequency range. At 1 MHz gain-bandwidth, you'll only get that gain somewhat below 200 Hz. Even a 1 mV input offset becomes 5 V after amplification by 5000.
R2 is also the impedance seen by the microphone after the input capacitor. You need this to be somewhat larger than the impedance of the microphone with its pullup and the input capacitor at the lowest frequency of interest. 10 Ω is way too small for that. 10 kΩ would be a better value.
Try two stages with a gain of 30 or so for starters and see where that gets you. That's a gain it can handle over reasonable frequencies with enough headroom left for the feedback to work. You also need to capacitively couple the two stages so that the input offset voltage does not accumulate thru all the stages.
Edit: Added circuit
I didn't have time to draw a circuit last night when I wrote the reply above. Here is a circuit that should do it:
This has a voltage gain of roughly 1000, which should be sufficient for a reasonable electret microphone. I may be a little too much, but its easy to add some attenuation.
The topology is rather different from your circuit. The most important single thing to note is that it doesn't try to produce the whole gain in one stage. Each stage has a gain of about 31. That leaves plenty of gain headroom at the maximum audio frequency of 20 kHz for the feedback, so the gain will be nicely predictable and flat accross the audio frequency range since the MCP6022 has a typical gain-bandwidth product of 10 MHz. The limiting factor will most likely be the microphone.
Unlike what I said before, the two stages do not need to be capacitively coupled to prevent the offset voltage accumulating along with the gain. That is because in this circuit, each stage has only a DC gain of 1, so the final offset is only twice the opamp offset. These opamps have only 500 µV offset, so the final offset is only 1 mV due to the opamps. There will be more due to the mismatch of R3 and R4. In any case, the output DC will be plenty close enough to 1/2 the supply to not eat into the A/D range in a meaningful way.
The DC gain of 1 per stage is achieved by capacitively coupling the feedback divider path to ground. The capacitor blocks DC, so each stage is just a unity follower for DC. The full AC gain is realized as the capacitor (C3 in the first stage) impedance gets small compared to the lower divider resistor (R7 in the first stage). This starts to happen at about 16 Hz. One drawback to this approach is that the time constant to settle is C3 times R7+R5, not just R7. This circuit will take a couple of seconds or so to stabalize after being turned on.
The microphone's output impedance is irrelevant to the choice of op-amp, because you "program" that aspect by a suitable op-amp circuit.
The low impedance of the mic means that the amplifier can have a low input impedance, in the thousands of ohms. But if the connection from the mic to the amplifier is short (we don't have to worry about stray capacitance of a cable), it doesn't have to. You can build the amplifier to have a relatively high input impedance, like 50 kOhms and up.
If you plan on using a coupling capacitor, like in the recommended circuit, a low input impedance will work against you: a low R means you will need a large C to maintain frequency response, which is linked to the RC product. (Since you give the audible range as 10 Hz (!) to 20 kHz, it can be assumed that you care about low frequency response).
The choice of op-amp depends on various parameters. This is a shopping question that is generally considered off-topic (on most stackexchange sites). You probably want it to be a low-noise unit suitable for audio, which has published distortion figures which are low. Then you have to consider your power supply: would a dual op-amp IC that drains up to 16 mA of current be acceptable? Or how about one that needs a minimum of 10V across its power rails to work properly: would that work? Cost: is it okay if the op-amp costs ten dollars? Or is fifty cents more appropriate? Output: does the op-amp have to produce output that goes almost all the way to the power rails? Or is it okay if it only goes to within a few volts of either rail before clipping? Manufacturing: are you comfortable with small, surface-mounted IC's, or would it be better to have a classic through-hole part with 0.1" pin spacing?
Whether or not a voltage divider is the best approach to power the mic depends on how much wattage will be wasted, and whether you can afford it.
Best Answer
Digital Signal Processors can do the FFT in real time. They have a particular architecture which means that certain instructions (MAC - multiply accumulate) can be done very quickly.
But then this turns in to an embedded software project too as well as the electronics design, if you're ok with that.
You can get development boards from companies such as Texas Instruments which contain a processor and break-out box arrangement for experimentation.