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The boundary of the mandelbrot set is the set of bifurcation points, each of which trace out a stable orbit which in turn can be seen as tracing out a polygon.
From the main minibulb clockwise the bulb orbits trace out a line, a triangle, square, pentagon etc.

The 3d version of the mandelbrot set surely would trace out polyhedrons rather than polygons.
A polyhedron surface is a 2d network or graph, this means each iteration produces two results instead of one, which means two alternate versions of the square function.
Angles of a polygon (complex numbers) would be replaced by 2d surface definitions.
The surface of a polyhedron is defined by two quaternions (q1,q2), which define how to rotate to move to the next face in two separate directions.
So the Julia set may be:
 (q1,q2) = (q1q1, q2q1) + (c1,c2)
 (q1,q2) = (q1q2, q2q2) + (c1,c2)
On the other hand, the square operation may need to use the Octonion square, so the Julia set would be:
 (q1,q2) = (q1,q2)^2 + (c1,c2)
 (q2,q1) = (q2,q1)^2 + (c2,c1)
A point is outside the set if any point in the binary tree escapes.

Either way, you could make some predictions about its mandelbrot set. It would be 8 dimensional, and would probably have the main bulb tracing out a plane, and possibly 3 bulbs that trace out tetrahedrons, and smaller bulbs for cube, octahedron etc. Some intermediate bulbs would trace out stellated polyhedra. The number of bulbs (stable repeating polyhedrons) would occupy a much smaller percent of the space than with the 2d mandelbrot. The 2d mandelbrot would probably exist inside this larger object as it traces out the polyhedrons where q2=identity, the prisms.

Now I have to work out how on earth I render an 8 dimensional julia set!

So the logic is surprisingly simple:

The mandelbrot fractal is its border
its border is a fractal of polygon cycles
a 3d equivalent would be a fractal of polyhedron networks
polyhedron networks are defined by two quaternions to represent two rotations of its 2d surface
the 2d equivalent of the single angle adding (angle + angle in z^2) is rotation convolution
Therefore you get
 (q1,q2) = (q1q1,q2q1) + (c1,c2)
 (q1,q2) = (q1q2,q2q2) + (c1,c2)
resulting in 8d julia sets, and a corresponding 8d mandelbrot set

This all seems almost inevitable to me. The problem is, like every other extended mandelbrot, it doesn't pass the universality test. You can translate or scale each iteration without changing the result, but can't rotate. This seems to be an inevitable consequence of using 3d rotations on a simple z^2 formula. A solution is to add extra rotations as a new free parameter, i.e.
 (q1,q2) = (Q1q1q1,Q2q2q1) + (c1,c2)
 (q1,q2) = (Q1q1q2,Q2q2q2) + (c1,c2)
but I doubt that a 16d mandelbrot set would be the simplest solution.... the problem is open.

One inevitable consequence of any such extension is that the iteration is very slow (intractable), so there would be no zooms to e280, lucky to get e10 with a powerful computer.

Perhaps you don't need a network in order to trace out polyhedrons, the quaternion mandelbrot doesn't do this because q^2 + c will tend towards rotating in the c axis, so will tend towards a 2d cycle. But it might be possible with:
 q = Q*q^2 + c
and its twin
 q = q^2*Q + c
since squaring a rotation and then rotating a little in a different direction seems at least capable of generating all the platonic solids (when c=0).
Q is a free parameter, so each mandelbrot is for a particular Q, much like it is a julia set itself.
rendering in 4d is not supported on any tools I know (even though it is perfectly possible) but that's ok because we can reduce c down to 3d by noting that any c can be rotated around the z axis so x=0 without any loss. We can also assume Q is unit length without loss of generality.

So this leads to a bounded set of 3d mandelbrots (one for each possible 3d orientation) which trace out polyhedrons, and it is universal in the sense that adding a quaternion at each iteration doesn't change the shape, scaling at iteration doesn't change the shape, and rotating by any quaternion at each iteration just moves to a different shape in the bounded set.

What will be interesting to see is how extruded it is... when Q=I it will be the normal quaternion mandelbrot, but for different Q we should see something more bumpy I think.

[edit] In hindsight this particular formulation cannot work, it just follows a cycle (not a true polyhedron) so can be modified in an extrusion way just as a prism can be modified to a fan/pyramid smoothly. i.e. the result will look extruded.