How can I calculate the amplitude of a signal from its power in dBm?
The Decibel Miliwatts Scale $dBm$ is the power ratio in Decibels, considering a reference of $P_0=1mW$.

$$P[dB]=10\text{log}_{10}(P/P_0)=10\text{log}_{10}(P[W]/0.001)$$

The general context is considering the signal amplitude $V$ in $volts$ and the power as an impedance load system $R$ in $\Omega$:

$$P=\frac{1}{R}V^2$$

Moreover, sinusoidal excitations are considered (for a given frequency also!), so we also can have:
$$P=\frac{1}{R}V_{rms}^2=\frac{1}{2}\frac{1}{R}V_{pk}^2$$

Thus:

$$
P[dBm] = 10\text{log}_{10}\left(\frac{V[V]^2}{R[\Omega]}\frac{1}{P_0}\right) = 20\text{log}_{10}\left(\frac{1}{\sqrt{R[\Omega]}}\frac{V[V]}{\sqrt{P_0}}\right)
$$

So, as reference: 


$1W$ equals $30dBm$, 
for a $50\Omega$ impedance system, this also equals $7.07V$, and 
under the sinusoidal case, this also equals $10V_{pk}$ and $7.07V_{rms}$.