Wow. I wonder who asked this and when... I guess the first question is: Prove that perfection is not possible. For regular tetrahedra that was done in 1958.

Next question -- perhaps open -- can this be done to get error under any positive epsilon? I and M Elgersma did this for regular tetrahedra a few years ago.

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@Ed, nice! This is so asking for 3D printing :-) like it was done with @Stan's tetrahedral chain challenge post. I could not resist generating GiF animation showing 3D view of the figure and the effect of offset constant. I added it to Ed's post. Here is the code:

frame=
ParallelTable[

offset=#+{0,0,Sqrt[1-1/4 Sec[Pi/8]^2]+4.05+2*Sin[3(u-Pi/2)]^2}&/@
(2PolyhedronData[{"Antiprism",4},"VertexCoordinates"]);

Graphics3D[Table[GraphicsComplex[RotationMatrix[k 2 Pi/13,{0,1,0}].#&/@offset,Polygon/@face],{k,0,12}],
Boxed->False,SphericalRegion->True,SphericalRegion->True,ImageSize->250,ViewPoint->7{.3,Sin[u],Cos[u/2]}],

{u,Pi/2,4 Pi+Pi/2,.1}];

Export["antiptism4.gif",frame]

Quite a few have asked ... "So how close is this?"

First, a slightly simpler antiprism:

root = Root[-1 + 2 #^4& , 2, 0];
v = {{1, -1, -root}, {-1, -1, -root}, {-1, 1, -root}, {1, 
    1, -root}, {Sqrt[2], 0, root}, {0, Sqrt[2], root}, {-Sqrt[2], 0, 
    root}, {0, -Sqrt[2], root}};
f = {{1, 2, 3, 4}, {5, 6, 7, 8}, {1, 8, 2}, {2, 7, 3}, {3, 6, 4}, {4, 
    5, 1}, {1, 5, 8}, {2, 8, 7}, {3, 7, 6}, {4, 6, 5}};
a4 = Polyhedron[v, f];

Next, a new function, PolyhedronFaceReflect, that refects the polyhedron on a given face:

PolyhedronFaceReflect[poly_, face_] := 
 Module[{v = poly[[1]], f = poly[[2]]},
  Polyhedron[
   RootReduce[
    ResourceFunction["ReflectPoints"][v[[Take[f[[face]], 3]]], v]], 
   f]]

With that, we can reflect on the faces and see the gap when the polyhedron is reflected precisely. A red sphere has been added to draw attention to the gap area.

dat = FoldList[PolyhedronFaceReflect, 
   a4, {3, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3, 5}];
Graphics3D[{{Red, Sphere[v[[6]] + {0, 0, .1}, .05]}, dat}, 
 Boxed -> False]

How far apart are the points in the gap triangles, for length 2 edges? 0.0268171 or 0.0189626

N[EuclideanDistance[dat[[1, 1, 6]], dat[[13, 1, 8]]]]
N[EuclideanDistance[dat[[1, 1, 3]], dat[[13, 1, 2]]]]

The gap is small enough that it's hard to draw attention to it. Here's a more twisted ring with a gap of 0.0330049:

dat = FoldList[PolyhedronFaceReflect, a4,{2,3,5,1,2,5,3,1,5,3,2,5,2,1,2,5,2,3,2}];
Graphics3D[{{Red,Sphere[v[[6]]+{0,0,.1},.05]},dat}, Boxed-> False]
N[EuclideanDistance[dat[[1, 1, 4]], dat[[20, 1, 4]]]]

Well, I took a look at the wild example, then took a wild guess that the following would be perfect:

dat = FoldList[PolyhedronFaceReflect, 
   a4, {3, 2, 5, 3, 2, 5, 2, 1, 2, 5, 2, 3, 5, 2, 3, 2, 1}];
Graphics3D[{{Red, Sphere[v[[6]] + {0, 0, .1}, .05]}, dat}, 
 Boxed -> False]

Let's try a check.

Table[EuclideanDistance[dat[[1, 1, k]], dat[[18, 1, k]]], {k, 5, 8}]

That returns {0, 0, 0, 0}, so it's a valid perfect toroid.
Can anyone make a 4-antiprism toroid with less than 18 antiprisms?

Ah: So that ends the story!! Back to the tetrahedron problem: That problem arose b/c it is quite simple to make perfect closed loops from cube, octahedron, dodecahedron, icosahedron. I have little models of those 4 here. So what you have discovered is that the same is true for your anti prism. But what is interesting is that it is nontrivial compared to the Platonic ones. Of course, maybe there is a simpler loop.... But finding what you found is great! Congrats. (Maybe even publishable. Oh wait: Why don't mwe make it an American Math Monthly problem. I am now the Proposal editor for the Amer Math Monthly, so I decide what we accept. This looks like a great problem! (Ie, True or False: there is an embedded chain of antiprisms) . Embedded means no intersections. One delicate point is that AMM Problems should not be published elsewhere. Maybe in a year or so you could remove the answer that is posted here.

Hmm... Found another ring of 18:

dat = FoldList[PolyhedronFaceReflect, a4,
   {3, 1, 2, 3, 5, 2, 1, 5, 2, 5, 1, 2, 5, 3, 2, 1, 3}];
Graphics3D[{dat, Red, dat[[1]]}, Boxed -> False]

(The "Red" lets me keep track of my starting point.)

Tom Sirgedas sent me a ring of 14 that's perfect.

dat = FoldList[PolyhedronFaceReflect, 
   a4, {1, 2, 1, 2, 10, 1, 8, 1, 2, 1, 2, 8, 1}];
Graphics3D[{{Red, Sphere[v[[6]] + {0, 0, .1}, .05]}, dat}, 
 Boxed -> False]

Then he lowered it down to 12.

dat = FoldList[PolyhedronFaceReflect, 
   a4, {10, 1, 8, 9, 1, 7, 8, 1, 10, 7, 1}];
Graphics3D[{{Red, Sphere[v[[6]] + {0, 0, .1}, .05]}, dat}, 
 Boxed -> False]