I know $H^1(E(C), Z_\ell)$ is isomorphic to $T_\ell(E(C))$ as group. I heard there are principle known as the lefschetz principle.
The legschetz principle says, roughly, that algebraic geometry over an arbitrary algebraically closed field of character $0$ is 'the same' as algebraic geometry over C.
Then, the titled theorem holds when we change C to Q's algebraic closure?
I saw also saw if we change C to Q, the isom holds (https://mathoverflow.net/questions/7318/etale-cohomology-and-l-adic-tate-modules).
Is this something related to lefschetz principle? ( At my stage, lefschetz principle seems very ambiguous, there are no concrete statement)
yes it is true this is because of two facts: first by "lefshetz principle" any elliptic curve has a model over some field $K_0\subset \mathbb{C}$ which means that for an elliptic curve $E/K$ there is $E_0/K_0$ such that $E=E_{K_0}\times_{K_0} K$(this is the precise statement of Lefschetz principle).In more down to earth terms, this means that there are only finitely many coefficients involved in defining an elliptic curve so your coefficients and hence your equations are already defined over some field of finite transcendence degree over $\mathbb{Q}$ and such a field can be embedded in $\mathbb{C}$.
the second fact does not have anything with lefchetz principle it is a general fact about etale cohomology which says that if $E/K_0$ is any variety then for any two algebraicly closed field $K,K'$ containing $K_0$, $H^*(E_K,\mathbb{Z}_l)=H^*(E_{K'},\mathbb{Z_l})$ you don't need that $Char(K)=0$ but you need that $l\not=Char(K)$