I guess it would be difficult to prove that the answero your question is "no", since proving that "no a priori reason exists" might be hard.
More modestly, I can say that I don't really know a good reason. However, here are some pointers:
One essential fact is that four of the five platonic solids are centrally symmetric; all but the tetrahedron. Croft observed: in an optimal inclusion two concentric polytopes will share a center.
Therefore, an optimal inclusion for two platonic solids, both of which are not the tetrahedron will result in a concentric situations. This makes these configurations considerably easier. This (at least hopefully somewhat) explains that the degree of the algebraic numbers in the 12 cases not involving the tetrahedron. Here is a table with the exact values:
Let's look at the cases involving the tetrahedron, that is the first row and the first columns of the table above.
The pairs $(T\subset C)$ and $(O\subset T)$ are again concentric, and so are the pairs $(T\subset D)$ and $(I\subset T)$.
The case $(T\subset O)$ is particulary simple to describe, since $T$ and $O$ share a face in the optimal solution.
What remains are really just $(C\subset T)$, $(D\subset C)$ and $(T\subset I)$. And indeed those are the one with solutions that are more algebraically involved. (The case $(C\subset T)$ not so much, since in the optimal configuration edges align and the constraints to begin with are "only" coming from $T$ and $C$ and not, as in the other two cases from the more complicated $D$ and $I$.)
to summarize: There are really only two cases where one might expect
more involved solutions, one of which is radical, the other not.
To address the second part of the question: "Or is there something wrong with the polynomial?": I hope not.
Let me give you a few more details, how one can calculate the polynomial in question.
We assume we know the incidences, i.e. what vertex of the tetrahedron lies on what face of the icosahedron and deduce the optimal solution from here.
The incidences are shown in the following picture: one vertex of T coincides with a vertex of I, one lies in an edges, the two remaining in two seperate faces of I.
I fix an icosahedron with edge-length $1$ and vertices $$\begin{align}
v_{0} & = \left(0,\,\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4}\right) \\
v_{1} & = \left(0,\,-\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4}\right) \\
v_{2} & = \left(\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0\right) \\
v_{3} & = \left(\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0\right) \\
v_{4} & = \left(\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0,\,\frac{1}{2}\right) \\
v_{5} & = \left(\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0,\,-\frac{1}{2}\right) \\
v_{6} & = \left(-\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0\right) \\
v_{7} & = \left(-\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0\right) \\
v_{8} & = \left(-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0,\,\frac{1}{2}\right) \\
v_{9} & = \left(0,\,\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4}\right) \\
v_{10} & = \left(0,\,-\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4}\right) \\
v_{11} & = \left(-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0,\,-\frac{1}{2}\right) \\
\end{align}$$
The four points of the tetrahedron are then given as follows:
$$\begin{align}p_0 =& v_{11}\\p_1 =& f_0v_0 + f_1v_2 + (1-f_1-f_0)v_6 \\ p_2 =& g_0v_1 + g_1v_3 + (1-g_0-g_1)v_7 \\p_3 =& e_0v_5 + (1-e_0)v_{10}\end{align}$$
For some positive variables $f_0, f_1, g_0, g_1$ and $e_0$.
For each $(i,j)\in\binom{[4]}{2}$, we have $(p_i-p_j)^2 = s^2$ for some real variable $s$.
Putting it all together, we obtain the following system of 6 equations with 6 variables:
$$\begin{align}
0 =& f_{0}^{2} + f_{0} f_{1} + f_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} f_{0} + \frac{1}{2} \, \sqrt{5} f_{1} - \frac{1}{2} \, f_{0} - \frac{1}{2} \, f_{1} + 1 \\
0 =& g_{0}^{2} + g_{0} g_{1} + g_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} g_{0} + \frac{1}{2} \, \sqrt{5} g_{1} - \frac{1}{2} \, g_{0} - \frac{1}{2} \, g_{1} + 1 \\
0 =& e_{0}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} e_{0} - \frac{1}{2} \, e_{0} + 1 \\
0 =& -\frac{1}{2} \, \sqrt{5} f_{0} g_{0} + f_{0}^{2} + f_{0} f_{1} + f_{1}^{2} - \frac{1}{2} \, f_{0} g_{0} - f_{1} g_{0} + g_{0}^{2} - f_{0} g_{1} - 2 \, f_{1} g_{1} + g_{0} g_{1} + g_{1}^{2} - s^{2} - f_{0} - g_{0} + \frac{1}{2} \, \sqrt{5} + \frac{3}{2} \\
0 =& -\frac{1}{2} \, \sqrt{5} e_{0} f_{1} + e_{0}^{2} - e_{0} f_{0} + f_{0}^{2} - \frac{1}{2} \, e_{0} f_{1} + f_{0} f_{1} + f_{1}^{2} - s^{2} - e_{0} - f_{1} + \frac{1}{2} \, \sqrt{5} + \frac{3}{2} \\
0 =& -\frac{1}{2} \, \sqrt{5} e_{0} g_{0} - \frac{1}{2} \, \sqrt{5} e_{0} g_{1} + e_{0}^{2} - \frac{1}{2} \, e_{0} g_{0} + g_{0}^{2} - \frac{1}{2} \, e_{0} g_{1} + g_{0} g_{1} + g_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} e_{0} + \frac{1}{2} \, \sqrt{5} g_{0} - \frac{1}{2} \, e_{0} - \frac{1}{2} \, g_{0} - g_{1} + 1 \\
\end{align}$$
This systems happens to have the solution
$$\begin{align}
f0=& 0.356785524577257 \text{..., zero of } 5751x^{16} + 54216x^{15} - 434286x^{14} - 466374x^{13} + 9452306x^{12} - 19323022x^{11} - 33022460x^{10} + 206565938x^{9} - 384738484x^{8} + 362774804x^{7} - 180708354x^{6} + 47907122x^{5} - 9497814x^{4} + 61518x^{3} + 895260x^{2} - 144318x + 401 \\
f1=& 0.352452740635196 \text{..., zero of } 5751x^{16} + 24732x^{15} - 16650x^{14} - 342012x^{13} - 901126x^{12} - 959712x^{11} + 743040x^{10} + 4747162x^{9} + 2877417x^{8} - 2704036x^{7} + 2767104x^{6} + 2220422x^{5} - 977942x^{4} - 2032764x^{3} - 504834x^{2} + 243738x + 64251 \\
g0=& 0.595049283356260 \text{..., zero of } 5751x^{16} + 46926x^{15} - 623124x^{14} + 3139674x^{13} - 7786156x^{12} + 2057864x^{11} + 49657348x^{10} - 189659290x^{9} + 400533046x^{8} - 535383644x^{7} + 467013828x^{6} - 298170008x^{5} + 167384576x^{4} - 74703516x^{3} + 17034188x^{2} - 698296x + 164 \\
g1=& 0.0729071548475811 \text{..., zero of } 5751x^{16} - 46224x^{15} - 194904x^{14} + 1789782x^{13} - 3522208x^{12} - 3388962x^{11} + 50328778x^{10} - 43454770x^{9} - 123966325x^{8} + 205675176x^{7} - 80109508x^{6} - 35043358x^{5} + 51387632x^{4} - 18740400x^{3} + 1053588x^{2} + 9072x - 369 \\
e0=& 0.645495309223693 \text{..., zero of } 71x^{16} + 972x^{15} - 2730x^{14} - 19898x^{13} + 64290x^{12} - 61466x^{11} + 46202x^{10} - 95276x^{9} + 136499x^{8} - 92310x^{7} + 51560x^{6} - 48088x^{5} + 35854x^{4} - 11920x^{3} - 2804x^{2} + 2406x - 41 \\
s=& 1.34744285033120 \text{..., zero of } 5041x^{32} - 1318386x^{30} + 60348584x^{28} - 924552262x^{26} + 5246771058x^{24} - 15736320636x^{22} + 29448527368x^{20} - 37805732980x^{18} + 35173457839x^{16} - 24298372458x^{14} + 12495147544x^{12} - 4717349124x^{10} + 1256858478x^{8} - 217962112x^{6} + 21904868x^{4} - 1536272x^{2} + 160801 \\
\end{align}$$ While the solution was found by using Newton's method combined with integer relations algorithms, the solution can be checked using exact calculations in $\mathbb{A}$ or in a smaller numberfield, which contains all the relevant numbers.
Indeed one can take the number field $F$ with defining polynomial any of the ones defining the solutions (substituting s by sqrt(s)).
They are all isomorphic and have discriminant $10637699079912558734361600000000 = 2^{16} \cdot 3^{4} \cdot 5^{8} \cdot 11 \cdot 466369383062945371$, divisible by the seemingly random prime.
