@AccidentalFourierTransform drew your attention to the Dirac operator, which, of course, can go Euclidean and extend to [all dimensions](https://en.wikipedia.org/wiki/Higher-dimensional_gamma_matrices). It has little to do with conformal FT, if that's where you want to go, because the conformal group is finite in all dimensions different than d=2. So disaggregate the conformal gig. 

First appreciate that Leon's construction can be superfluously replicated to 2×2 matrices (Pauli, duh!), through defining the Euclidean 2d Dirac operator, 
$$
D=\sigma _x\partial_x + \sigma_y \partial_y =    \begin{pmatrix}
      0&\partial_x -i\partial_y\\
      \partial_x+i\partial_y&0
    \end{pmatrix} ~~~\leadsto \\
D^2 = \Delta ~~{\mathbb 1}_2. 
$$
So far superfluous, but you may extend this to 3d, again with 2×2 matrices since their spinors are the same for an even dimension and its odd-one-higher one,
$$
D=\sigma _x\partial_x + \sigma_y \partial_y + \sigma_z \partial_z =    \begin{pmatrix}
      \partial_z&\partial_x -i\partial_y\\
      \partial_x+i\partial_y&-\partial_z
    \end{pmatrix} ~~~\leadsto \\
D^2 = \Delta ~~{\mathbb 1}_3.
$$

And so on. The link provided will let you write down the Dirac gamma matrices in ***all*** dimensions, rectify your Minkowski metric by multiplying the spacelike ones with *i*, and dot them with the d-dimensional gradient to factorize your Laplacian in *all* dimensions, as above. Again, I'm not clear what you imagine you can get for CFTs in all dimensions this way. (To me the gig looks useless, but I should not aim to discourage creative thinking...)