Assume that the Tate-Shafarevich group $\operatorname{Sha}(K,E)$ is finite. For each element of $\operatorname{Sha}(K,E)$, take the corresponding genus $1$ curve, and let $V$ be the product of all of them. Then the field version of your first question is exactly asking for a field extension $L$ such that $V(L)$ is nonempty. As you say, there are many such $L$, but is there a canonical such field $L$? Yes, let $L$ be the function field of $V$! (Admittedly, this is rather tautological, but I'm not sure that one can do much better if you really want a canonical answer.)

As for the "norm map", it is perhaps more commonly known as the corestriction map in Galois cohomology, $\operatorname{Cores} \colon H^1(L,E) \to H^1(K,E)$, which can be restricted to the locally trivial elements if you want. As surely you know already, the fact that $\operatorname{Cores} \circ \operatorname{Res}$ equals multiplication by $[L:K]$ is responsible for the generalization of what you called Heegner's lemma, that if an element of order $n$ in $H^1(K,E)$ restricts to $0$ in $H^1(L,E)$ for some finite extension $L$ of $K$, then $n$ divides $[L:K]$.

What I find more amusing is the following: Let's continue to assume finiteness of Tate-Shafarevich groups. Let $X$ be a genus $1$ curve corresponding to $x \in \operatorname{Sha}(K,E)$ of prime order $n$; then there is some $y \in \operatorname{Sha}(K,E)$ such that the Cassels-Tate pairing $\langle x,y \rangle$ gives $1/n$ in $\mathbf{Q}/\mathbf{Z}$. For every degree-$n$ extension $L$ of $K$ such that $X(L) = \emptyset$ (i.e., the restriction $x_L:=\operatorname{Res}_{L/K}(x)$ is nonzero), there is a new element $z \in \operatorname{Sha}(L,E)$ such that the Cassels-Tate pairing $\langle x_L,z \rangle$ for $L$ gives $1/n$: this $z$ has to be new, i.e., not just $y_L$, because of how the Cassels-Tate pairing behaves under field extension: $\langle x_L, y_L \rangle = [L:K] \langle x,y \rangle = 0$. So we expect to get elements of $\operatorname{Sha}(L,E)$ for various $L$ popping up all over the place!

EDIT: This is a completely new answer.

I will prove that your specific suggestion of defining a Hilbert class field of an elliptic curve $E$ over $K$ does not work. I am referring to your proposal to take the smallest field $L$ such that the corestriction (norm) map $\operatorname{Sha}(L) \to \operatorname{Sha}(K)$ is the zero map. (I have to assume the Birch and Swinnerton-Dyer conjecture (BSD), though, for a few particular elliptic curves over $\mathbf{Q}$.)

Theorem: Assume BSD. There exists a number field $K$ and an elliptic curve $E$ over $K$ such that there is no smallest field extension $L$ of $K$ such that $\operatorname{Cores} \colon \operatorname{Sha}(L,E) \to \operatorname{Sha}(K,E)$ is the zero map.

Proof: We will use BSD data (rank and order of Sha) from Cremona's tables. Let $K=\mathbf{Q}$, and let $E$ be the elliptic curve 571A1, with Weierstrass equation $$y^2 + y = x^3 - x^2 - 929 x - 10595.$$ Then $\operatorname{rk} E(\mathbf{Q})=0$ and $\#\operatorname{Sha}(\mathbf{Q},E)=4$. Let $L_1 = \mathbf{Q}(\sqrt{-1})$ and $L_2 =\mathbf{Q}(\sqrt{-11})$. It will suffice to show that the Tate-Shafarevich groups $\operatorname{Sha}(L_i,E)$ are trivial.

Let $E_i$ be the $L_i/\mathbf{Q}$-twist of $E$. MAGMA confirms that $E_1$ is curve 9136C1 and $E_2$ is curve 69091A1. According to Cremona's tables, $\operatorname{rk} E_i(\mathbf{Q})=2$ and $\operatorname{Sha}(\mathbf{Q},E_i)=0$, assuming BSD. Thus $\operatorname{rk} E(L_i) = 0+2=2$ and $\operatorname{Sha}(L_i,E)$ is a $2$-group. On the other hand, MAGMA shows that the $2$-Selmer group of $E_{L_i}$ is $(\mathbf{Z}/2\mathbf{Z})^2$. Thus $\operatorname{Sha}(L_i,E)[2]=0$, so $\operatorname{Sha}(L_i,E)=0$.