Magnitude of the Gradient in Frequency Domain
Remember that as part of computing the gradients of an image, you are really just convolving your image, with (two) spatial filters. This then gives you two 'gradient' images, which you can then combine.

For example, to compute the x-direction-gradient-image $G_x$, you would convolve your image with a filter $f_x$ that might look like this $\begin{bmatrix} 1 &amp; 0 &amp; -1\end{bmatrix}$. Similarly, to attain the y-direction-gradient-image,  $G_y$, you might convolve your image with a filter $f_y$ that looks like this $\begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}$. 

Then, your final gradient image would be $G = \sqrt{G_x^2 + G_y^2}$.

So far so good. However, you don't need to do the convolution pixel by pixel. Instead, you could simply take the 2D DFT of $f_x$, multiply it pixel by pixel with the 2D DFT of the image, and then take the 2D IDFT back into the spatial domain. This would then also give you $G_x$. (Do the same to find $G_y$). 

So the answer is at least a partial yes, in that you can compute the x and y gradient images via the FFT algorithm, (an implementation of the DFT), and then simply treating each of them as an element in a vector, and finding the L2 norm.