(This may have errors - I'm not an algebraic number theorist.)

We have a complete description of the multiplicative structure of $F = \mathbb{Q}(i)$. It is:
$$\mathbb{Q}(i)^\times \cong \left( \bigoplus_{p \cong 1 \mod 4} (\mathbb{Z} \oplus \mathbb{Z}) \right) \oplus \left( \bigoplus_{\text{other } p} \mathbb{Z} \right) \oplus \mathbb{Z}/4\mathbb{Z}.$$
Note that there are no $n$-divisible subgroups, except the 4th roots of unity (when $n$ is odd). This is a good sign of infinite degree.

Each prime $p$ congruent to 1 mod 4 can be written as product of primes $(a+ib)(a-ib)$, with $a$ and $b$ unique up to obvious symmetries. We find that $\frac{a+ib}{a-ib} \in T$, and is a primitive element in the copy of $\mathbb{Z} \oplus \mathbb{Z}$ in the big sum corresponding to $p$. In particular, it is not an $n$th power for $n \geq 2$.

We can now construct a sequence of fields $F=F_0 \subset F_1 \subset \dots$, where $F_k$ is given by starting with $F_{k-1}$, and adjoining an $n$th root of the number $\frac{a+ib}{a-ib}$ corresponding to some prime congruent to 1 mod 4 over which $F_{k-1}$ is unramified. Since finite extensions are ramified over finitely many primes, and adjoining the $n$th root creates ramification over $p$, we have strict containment at each step, and the chain does not terminate after finitely many steps. The union of the chain is an infinite degree extension that is contained in $E$, so $E$ has infinite degree over $F$.