Calculating the cutoff frequency of a highpass digital filter
Concerning the cut-off frequency, there's really not much of a difference between analog and digital filters. $3\textrm{ dB}$ is common, but any other value is fine, as long as you specify it and people know what you're talking about.

In the case of the given filter, it's quite straightforward to compute any cut-off frequency. Let's choose the $3\textrm{ dB}$ cut-off frequency:

$$H(e^{j\omega})=1-\frac12\left(e^{j\omega}+e^{-j\omega}\right)=1-\cos(\omega)\tag{1}$$

Since $H(e^{j\omega})$ is real-valued and since $H(e^{j\omega})\ge 0$, we can simply solve the equation

$$1-\cos(\omega_c)=\frac{2}{\sqrt{2}}=\sqrt{2}\tag{2}$$

Note that the maximum of $H(e^{j\omega})$ is $2$, so the value of $H(e^{j\omega})$ at the $3\textrm{ dB}$ cut-off frequency $\omega_c$ is $\sqrt{2}$.

From $(2)$ we get $\omega_c\approx 1.9979$. The actual cut-off frequency in Hertz is given by

$$f_c=\frac{\omega_c}{2\pi}f_s\tag{3}$$

where $f_s$ is the sampling frequency.