I don't know anything about hypergeometric functions, so this is not a direct answer to your question. But, I have thought a lot about the problem of detecting complex multiplication of elliptic curves (and certain higher-dimensional analogues for abelian varieties).
Suppose you are given an algebraic integer $j$, and you wish to know whether it is a CM j-invariant. Then there is a sort of night-and-day algorithm you can perform here, where by night you reduce the elliptic curve modulo various primes and keep in mind the fact that if your elliptic curve has CM, then half the time (in the sense of density) you will get a supersingular elliptic curve and the other half you will get a CM elliptic curve whose characteristic p endomorphism algebra is the same as the algebra you started with. Thus, in practice, if your curve does not have CM, you will fairly quickly be able to rule it out by finding two primes of ordinary reduction with different endomorphism algebras. So far this is just a probabilistic algorithm. Once you figure out that the CM field is either a particular quadratic field K or there is no CM at all, you compute (e.g. by classical CM theory as described e.g. in Cox's book Primes of the form...) you compute the $j$-invariants of elliptic curves with K-CM. There are infinitely many of these, because the j-invariant depends on the endomorphism ring (equivalently, the conductor of the order), but you can either just compute all of them in order of conductor or look more carefully at the mod p reductions and get a bound on what the conductor could be.
[Edit: Actually, you can figure out exactly what the CM order must be by computing the endomorphism ring at any two primes of ordinary reduction. It is a theorem that if $E$ is a curve with CM by the order of conductor $f$ in a CM field $K$ and $p$ is a prime of ordinary reduction, the conductor of the reduced endomorphism ring is $f/p^{ord_p(f)}$, i.e., you just strip away the $p$-part of the conductor.]
This is not the state of the art, though. Rather, see the paper
Achter, Jeffrey D.
Detecting complex multiplication. (English summary) Computational aspects of algebraic curves, 38--50,
Lecture Notes Ser. Comput., 13, World Sci. Publ., Hackensack, NJ, 2005.
[A copy is available via his webpage http://www.math.colostate.edu/~achter/.]
In the paper, Achter uses Faltings' theorem and the effective Cebotarev density theorem to eliminate the "day" part of the algorithm. He also gives a complexity analysis and explains why this is faster than what I sketched above.
Finally, I'm sure the questioner knows this, but others may not: for elliptic curves over $\mathbb{Q}$ there's no need to do any of this. Rather you just compute the $j$-invariant and see whether it's one of the $13$ $j$-invariants of CM elliptic curves over $\mathbb{Q}$ associated to the $13$ class number one quadratic orders (yes, this relies on the Heegner-Baker-Stark resolution of Gauss' class number one problem). For the list, see e.g.
modular.fas.harvard.edu/Tables/cmj.html
Here is a case where it is non-Abelian. I use $K$ of class number 3. If I use the Gross curve, it is Abelian. If I twist in $Q(\sqrt{-15})$, it is Abelian for every one I tried, maybe because it is one class per genus. My comments are not from an expert.
> K<s>:=QuadraticField(-23);
> jinv:=jInvariant((1+Sqrt(RealField(200)!-23))/2);
> jrel:=PowerRelation(jinv,3 : Al:="LLL");
> Kj<j>:=ext<K|jrel>;
> E:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]);
> HasComplexMultiplication(E);
true -23
> c4, c6 := Explode(cInvariants(E)); // random twist with this j
> f:=Polynomial([-c6/864,-c4/48,0,1]);
> poly:=DivisionPolynomial(E,3); // Linear x Linear x Quadratic
> R:=Roots(poly);
> Kj2:=ext<Kj|Polynomial([-Evaluate(f,R[1][1]),0,1])>;
> KK:=ext<Kj2|Polynomial([-Evaluate(f,R[2][1]),0,1])>;
> assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here
> f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1];
> GaloisGroup(f); /* not immediate to compute */
Permutation group acting on a set of cardinality 12
Order = 48 = 2^4 * 3
> IsAbelian($1);
false
This group has $A_4$ and $Z_2^4$ as normal subgroups, but I don't know it's name if any.
PS. 5-torsion is too long to compute most often.
Best Answer
Under your setting \begin{equation} \operatorname{Sel}(E/K_{\infty})\simeq\operatorname{Sel}(E/\mathbb Q_{\infty})\oplus\operatorname{Sel}(E\otimes\chi/\mathbb Q_{\infty}) \end{equation} where $\chi$ is the quadratic character attached to $K/\mathbb Q$ and all these $\mathbb Z_{p}[[X]]$-modules are torsion. Hence \begin{equation} \mu(\operatorname{Sel}(E/K_{\infty}))=\mu(\operatorname{Sel}(E/\mathbb Q_{\infty}))+\mu(\operatorname{Sel}(E\otimes\chi/\mathbb Q_{\infty})). \end{equation} Indeed, if $\mu(\operatorname{Sel}(E/K_{\infty}))=0$ then $\mu(\operatorname{Sel}(E/\mathbb Q_{\infty}))=0$.
However, the very same argument suggests that it is probably very hard to show that $\mu(\operatorname{Sel}(E/K_{\infty}))=0$ without showing that $\mu(\operatorname{Sel}(E/\mathbb Q_{\infty}))=0$.
What has been known since the works Gillard, Schneps, Hida, Hsieh... is that it is sometimes possible to prove that $\mu$-invariant vanishes over the $\mathbb Z_p^2$-extension $L_\infty/K$ (the composite of the cyclotomic and anticyclotomic extension of $K$). Unfortunately, in that case, there is no obvious way (and indeed, at present no known way) to deduce from those results the vanishing of the $\mu$-invariant over $\mathbb Q_\infty$, as the specialization of a power-series in $\mathbb Z_p[[X,Y]]$ with vanishing $\mu$-invariant at $Y=0$ can very well have non-vanishing $\mu$-invariant.