Normalization factor in the convolution theorem
The choice of the normalization factor is just a matter of convention. Note that the specific correspondence between convolution in the time domain and multiplication in the frequency domain with a scaling of $\sqrt{2\pi}$, as shown in your question, applies only to the unitary definition of the Fourier transform with angular frequency as the independent variable in the frequency domain:
$$F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-j\omega t}d\omega\tag{1}$$
If we use the non-unitary definition (as is common in the field of signal processing)
$$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}d\omega\tag{2}$$
then the correspondence is simply
$$\mathcal{F}\{(f\star g)(t)\}=F(\omega)G(\omega)\tag{3}$$
The same is true if we use frequency instead of angular frequency.
For the discrete-time Fourier transform (DTFT), we have exactly the same choices.
For the discrete Fourier transform (DFT), where we deal with samples in both domains, the same holds for the normalization factor $N$, where $N$ is the DFT length. In the field of signal processing, the DFT is commonly defined as a non-unitary transform:
$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{4}$$
The inverse transform is then given by
$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{5}$$
The unitary DFT definition is
$$X[k]=\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{6}$$
and the corresponding inverse transform is
$$x[n]=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{7}$$