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Since your horizontal axis unit is just i and the vertical unit is (cos(pi/8),sin(pi/8)), I suspect that you'd get a severely skewed version of the standard Mandelbrot set.  Likewise with your second question, depending, of course, on how you define the 3d math.

I want to second lkmitch: ⁴√(-1) is just √(2)/2 + √(2)/2 i = cos(pi/4) + sin(pi/4) i.  (lkmitch accidentally said pi/8.)  (Technically, there are four complex numbers z which satisfy z*z*z*z = -1.  Any of those could be called the "fourth root of -1".  Roots in complex numbers are a bit weird.)

In the 2D case,
a √(-1) + b ⁴√(-1) = (a + b√(2)/2) + b√(2)/2 i.
And lkmitch, is right:  This change of basis would give you a picture of the Mandelbrot set which is rotated and skewed.

As for drawing the Mandelbrot Set in this axis system, it would be easy to do.  Learn how to draw the Mandelbrot set in your favorite program language and then modify the program slightly.  If you are new to programming, this would be a fun project to get started.

As for the 3D version,  I assume you want the point (a, b, c) to correspond to the complex number
a + b √(-1) + c ⁴√(-1) = (a + c √(2)/2) + (b + c √(2)/2) i.
Notice every complex number is already in the xy plane (where c = 0).  Therefore, on the xy plane there will be a picture of the Mandelbrot set and then that same picture will also appear in every plane parallel to the xy plane (except it will be shifted so that center of the picture is in another location).