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I have seen a few things in this forum looking something like what I have in mind, but only quite. What I have in mind is 3D-Mandelbrot which, when sliced at z=0 actually looks like the 2D-Mandelbrot.
here my attempt:

expand the rules for multiplication and summation of complex numbers to the 3rd dimension:
Z(n) = Z(n-1)^2 + C
would then mean convert Z(n-1) from Cartesian to spherical coordinates multiply the lengths and add the angles, convert it back to Cartesian and add C

I have done 256 iterations but only on a 154x154x154 array. here the results.

an isosurface on points which are finite:

count how many iterations it takes for vector length to exceed 4. plot a isosurface for each iteration 256 to 1 in steps of 4. each isosurface with a alpha of 0.1:
   
tell me what you think...

  Yeah, an oldie but a goodie.

  It's the equivalent of taking:

the complex number z1=  sqrt(start_x^2 +start_y^2) + i start_z
  real_z1=  real part of z1
  imag_z1= imaginary part of z1

the complex number z2= start_x + i start_y
  real_z2=  real part of z2
  imag_z2= imaginary part of z2

magnitude of z2,  r =  sqrt (start_x^2 +start_y^2)

z1= z1 ^ n
z2= z2 ^ n
r=    r ^ -n

new_x = real_z1 * real_z2 *r
new_y = real_z1 * imag_z2 *r
new_z = imag_z1

  add in your pixel values appropriately, so on and so forth....

  However, the really detailed one is the "opposite" complex triplex:

 you take:
z1 = start_x + i sqrt(start_y^2+ start_z^2)...
etc...

  You don't get a 2d mandelbrot shape, but then again you shouldn't expect one due to the mathematics involved (as the 2d brot is skewed over the y axis, you get a double skew due to the 2 imaginary components for the 3d version, rather than a simple single skew)....

So do I - another newbie's attempt:

z=a+bi+cj
z²=(a²-2bc) + i (2ab-c²) + j (2ac-b²)
with a special multiplication:
ii=-j jj=-i ij=ji=-1
e=1 ei=ie=i ej=je=j
http://members.aon.at/evab/
(german, but less text, more images)