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4, 6, 8 triangles can make a tetrahedron and up.
6, 8, 9, 10 quadrilaterals can make a cube and up.
12, 16, 18, 20 pentagons can make a tetartoid or dodecahedron and up.
7, 8, 9, 10 hexagons can make make a Szilassi toroid and up.
12, 24 heptagons can make a heptagonal dodecahedron or Klein quartic 3-torus (shown below).

klein quartic

4, 6, 12, 7, 12, ... what is next in this sequence?

Could it possibly be 15, with Foster graph F040, which I call the Moving Day graph after Loyd's Moving Day puzzle? If so, how can those octagons be made planar to contain a 3d-printable space? Or is it some other graph? Perhaps 24 octagons can be linked together in a way similar to the 24-cell?

Moving Day Graph

For nonagons, is it Foster F060A, or the Biggs-Smith graph?

Anything between 12 and 24 for heptagons?

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I believe I have an answer to one of your many sub-questions, and here it goes:

No, the minimum for octagons is neither 15 nor 24.

Minimal octagons

This guy has 12. Each face is one of the following shapes:

faces

The geometric symmetry is relatively low ($\bar4$ in Hermann–Mauguin notation).

The construction details are below:

Take this one (left), overlap is with a tetrahedron (right):

construction-1

I mean, like this:

construction-2

Then subtract the tetrahedron from the other one.

construction-3

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Going against the customs of this site and my own habits, I decided to add another answer to the same question, because I can't think of a better place where to put this. Strictly speaking, it is an answer to a different question, which the OP did not ask, but probably had in mind.

Behold the figure made of twelve 11-gons. Undecagons figure

Pity it has many pairs of faces share more than one edge, and consequently many pairs of faces not sharing an edge, and hence is not the much sought-after hypothetical continuation to the series of polyhedra in which every face meets every other (tetrahedron $\to$ Szilassi $\to\;???$).

Each face is one of the following shapes:

faces

The geometric symmetry, believe it or not, is again $\bar4$.

The construction details are below:

First we take a prism made out of a concave quadrilateral. beam

Then we add another such beam in a criss-cross fashion, so that their sharp edges cut into each other a bit.

two beams

And finally we cut a hole of this shape in a tetrahedron.

still view

Topologically this is a sphere with six handles, just as it should be.

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