Your electret's datasheet specifies:
It's been a while, but this is an old way of writing out sensitivity.
One bar is \$10^5\:\text{Pa}\$. So \$1\:\mu\text{bar}\$ is \$100\:\text{mPa}\$.
Today, we usually use one Pascal as the reference, so this means we need to add \$20\:\text{dB}\$ to the written specification for comparison with more common specs today.
\$60\:{\text{dB}_\text{SPL}}\$ is a loud conversational level. \$40\:{\text{dB}_\text{SPL}}\$ is the lower end of conversational levels.
Loud, the output would be \$20\:\mu\text{Pa} \times 10^{^\frac{60\:{\text{dB}_\text{SPL}}-54\:{\text{dB}_\text{SPL}}+20\:{\text{dB}_\text{SPL}}}{20}} \$ or about \$400\:{\mu\text{V}_\text{PK}}\$. It will be about 10 times less, or about \$40\:{\mu\text{V}_\text{PK}}\$ at the lower end of conversational levels.
This assumes, as stated, that \$R_\text{L}=1\:\text{k}\Omega\$. You can use a higher voltage where possible, but this also means a larger \$R_\text{L}\$. In this case, I'd just recommend a standard \$9\:\text{V}\$ alkaline battery as the voltage source. I usually anticipate that the electret will require about \$1.5\:\text{V}\$ itself. And since a standard \$9\:\text{V}\$ alkaline battery is really more like \$8.5\:\text{V}\$, I'd use \$7\:\text{V}\$ as the resistor drop. I also usually expect an electret requires \$500\:\mu\text{A}\$. So this means \$R_{_\text{L}}=\frac{8.5\:\text{V}-1.5\:\text{V}}{500\:\mu\text{A}}=14\:\text{k}\Omega\$.
I'd select \$10\:\text{k}\Omega\$, though. (Some of the electrets can use a little more current.)
Given this resistor value is 10 times the one they specified, you can expect about 10 times the AC output voltage, but obviously now with a source resistance that is also 10 times higher. So between \$400\:{\mu\text{V}_\text{PK}}\$ and \$4\:{\text{mV}_\text{PK}}\$ and with a source impedance of \$10\:\text{k}\Omega\$.
To get that up into the range of \$1\:{\text{V}_\text{RMS}}\$ would require a final voltage gain (after attentuations are accounted) of \$20\cdot\log_{10}\left(\frac{1\:{\text{V}_\text{RMS}}}{4\:{\text{mV}_\text{PK}}\,\cdot\, \frac1{\sqrt{2}}}\right)\approx 51\:\text{dB}\$ (about 350 times.)
But that's where I'd start the design process.
The first stage should be carefully designed to use low-noise transistors. (Sufficiently low-noise opamps are power-hogs, boutique, and otherwise not needed -- so BJT-only.) The voltage gain should be very modest -- perhaps a factor of 3 or 4 (\$+9.5\:\text{dB}\$ to \$+12\:\text{dB}\$.) But no more than about \$+15\:\text{dB}\$. A class-A stage should be fine, but use a current source as the collector load for it. Once that stage is done, then the 2nd stage can be designed to provide the remaining gain of around \$+39\:\text{dB}\$. (And that can be a decent audio amp.)
Regardless, this will require a minimum of two stages of amplification: the pre-amplifier (described above) and an amplifier stage that provides necessary remaining voltage gain and matching up with the ADC input for the MCU. (We've not yet discussed the MCU ADC.) But 3 stages would not be off the table.
Not doing the necessary analysis and then design and construction steps to get from A to B is likely why you may be having troubles.
Example of first stage design
There is a design process for a similar amplifier found here at the EESE site that includes the following schematic:
This is about how I would approach a first stage design using low noise BJTs (like the BC549 and BC559), except that the above design isn't just a first stage, wasn't designed for low noise, and is instead designed to directly drive a speaker.
There's a comment there, from G36, saying that the above design is similar to one by John Linsley Hood, dating back to this 1969 article (now preserved at Elliot's site).
This brings me to the point. There is an excellent pair of articles at Elliot Sound Products:
That 2nd article (see Figure 4 there) provides a design extremely similar to what I was thinking about, above. In fact, it also uses the same \$9\:\text{V}\$ battery and specifies the same \$10\:\text{k}\Omega\$ resistor!
(Once you decide on the battery, the resistor mostly just falls out from that. So we reached the same place likely for similar reasons.)
Elliot's design also selects a current source driven class-A design (which makes more sense that going for class-AB, since accuracy is important and power isn't the issue.)
We are essentially on a similar page, even down to the expected gain region. I would guess that any other good quality design will fit this pattern, too.
I've redrawn it in a way I prefer for understanding it. And I've replaced his output capacitors as this will be feeding into the next stage and I just wanted to test it. Click on the image to make it larger and more readable:
I've also included an electret simulation of my own in the lower left corner, for testing purposes. (And the obvious \$9\:\text{V}\$ battery source simulation, as well.)
The top graph shows the loaded microphone output, which is about \$2.5\:{\text{mV}_\text{RMS}}\$. Unloaded, this would be closer to \$2.8\:{\text{mV}_\text{RMS}}\$. But the amplifier will load it down a little.
The bottom graph shows the unloaded output of the amplifier stage. (I used a \$10\:\text{M}\Omega\$ resistor, which is essentially no load.) This is about \$10.6\:{\text{mV}_\text{RMS}}\$. So the gain is about \$+12.5\:\text{dB}\$.
I'd probably add a \$47\:\text{pF}\$ capacitor between \$Q_4\$'s collector and base to roll off the high end. But that's all I'd probably do.
I think this confirms the advice I gave you, earlier.
You imagined that there was some kind of magic stuff inside of an electret that makes it directly compatible with an MCU ADC. But this just is not the case.
