What is an RC time constant and why do I have to wait for the capacitor to carge up before taking samples with ADC?
The time constant of an RC element is calculated \$ \tau = R \cdot C \$.  

Example: With \$R = 10 k\Omega \$ and \$C=10nF\$ this gives you \$\tau = 10k\Omega \cdot 10nF = 100µs\$.
If you have to wait for 11 RC time constants you have to wait for \$11 \cdot \tau\$, in the example above this would be 1.1ms.

Details:
\$\tau\$ corresponds to the time it would take a capacitor to charge up to the applied voltage over a resistor R, if the initial charging speed (at t=0) would charge the whole capacitor. As you probably know, the charging speed decreased while charging, because the voltage drop over the resistor is decreasing. So after 1 \$\tau \$ the capacitor is only charged to 63 %. 


Source: https://de.wikipedia.org/wiki/RC-Glied

In typical applications it is assumed that a capacitor is charged up to full voltage after \$5\tau\$ (> 99%), but 99% is not enough if you want to measure with 16 Bit resolution:

$$1 - \frac{1}{2^{16}} = 1 - \frac{1}{65536} = 1 - 0.00001526... = 0.99998474...$$  

So the input capacitance has to be charged up to 99.99847%. With a charging of the capacitor with

$$ U_{(t)} = U_{max} \cdot (1-e^{-\frac{t}{\tau}}) $$

this is reached after

$$ \frac{U_{(t)}}{U_{max}} = 0.99998474 = (1-e^{-\frac{t}{\tau}})$$
$$ e^{-\frac{t}{\tau}} = 0.00001256 $$
$$ {\frac{t}{\tau}} = 11.09$$
$$ t = 11.09 \tau = 11.09 \cdot R\cdot C$$

As you can see, it takes the capacitor little over 11 RC time constants to charge to a high enough voltage to measure with 16 bit accuracy.