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To partly answer your last two questions: It's not too hard to write down a sequence of abelian varieties $A_i/\mathbf{Q}$ such that $\mathrm{rank}A_i(\mathbf{Q})=0$ but $\mathrm{rank}A_i(\mathbf{Q}(\sqrt{-1}))\to \infty$ as $i\to\infty$. More precisely, if $p\equiv 1\ \mathrm{mod}\ 4$ is large enough, there is some newform $f$ of weight $2$ on $\Gamma_0(p^3)$ such that

a. $L(s,f)$ has root number $+1$, and $L(1/2,f)\neq0$, which in fact implies $L(1/2,f^{\sigma})\neq0$ for all Galois conjugates of $f$ by some work of Shimura;

b. $L(s,f\otimes \chi_{-4})$ has root number $-1$ (which is automatic from a. and my choice of $p$) and $L'(1/2,f \otimes \chi_{-4})\neq 0$, which again implies $L'(1/2,f^{\sigma} \otimes \chi_{-4})\neq 0$ for all Galois conjugates of $f$ by the Gross-Zagier formula and its generalizations.

The existence of such a newform follows from the asymptotic evaluation

$\sum_{f\in S_2^{new}(\Gamma_0(p^3)),\ \varepsilon(f)=+1}L(1/2,f)L'(1/2,f\otimes \chi_{-4}) \sim cp^3$

for some real $c>0$, which nowadays is a standard application of the Petersson formula (see e.g. the final chapter of Iwaniec and Kowalski's book). Given such an $f$, let $A_f$ be the corresponding optimal quotient of $J_0(p^3)$. Then a. guarantees that $\mathrm{rank}A_f(\mathbf{Q})=0$ by the work of Kolyvagin/Zhang/Longo/Tian-ZhangKolyvagin-Logachev/Zhang/Longo/Tian-Zhang, while b. guarantees that $\mathrm{rank}A_f(\mathbf{Q}(\sqrt{-1}))=\mathrm{dim}A_f$ by the same group of people, and $\mathrm{dim}A_f \geq \frac{p-1}{2}$ (see e.g. this question), so we're done.

This whole argument works with $\mathbf{Q}(\sqrt{-1})$ replaced by any fixed imaginary quadratic field. If you demand the same dichotomy for a sequence of $A$'s of bounded dimension, I imagine it's not known.

1

To partly answer your last two questions: It's not too hard to write down a sequence of abelian varieties $A_i/\mathbf{Q}$ such that $\mathrm{rank}A_i(\mathbf{Q})=0$ but $\mathrm{rank}A_i(\mathbf{Q}(\sqrt{-1}))\to \infty$ as $i\to\infty$. More precisely, if $p\equiv 1\ \mathrm{mod}\ 4$ is large enough, there is some newform $f$ of weight $2$ on $\Gamma_0(p^3)$ such that

a. $L(s,f)$ has root number $+1$, and $L(1/2,f)\neq0$, which in fact implies $L(1/2,f^{\sigma})\neq0$ for all Galois conjugates of $f$ by some work of Shimura;

b. $L(s,f\otimes \chi_{-4})$ has root number $-1$ (which is automatic from a. and my choice of $p$) and $L'(1/2,f \otimes \chi_{-4})\neq 0$, which again implies $L'(1/2,f^{\sigma} \otimes \chi_{-4})\neq 0$ for all Galois conjugates of $f$ by the Gross-Zagier formula and its generalizations.

The existence of such a newform follows from the asymptotic evaluation

$\sum_{f\in S_2^{new}(\Gamma_0(p^3)),\ \varepsilon(f)=+1}L(1/2,f)L'(1/2,f\otimes \chi_{-4}) \sim cp^3$

for some real $c>0$, which nowadays is a standard application of the Petersson formula (see e.g. the final chapter of Iwaniec and Kowalski's book). Given such an $f$, let $A_f$ be the corresponding optimal quotient of $J_0(p^3)$. Then a. guarantees that $\mathrm{rank}A_f(\mathbf{Q})=0$ by the work of Kolyvagin/Zhang/Longo/Tian-Zhang, while b. guarantees that $\mathrm{rank}A_f(\mathbf{Q}(\sqrt{-1}))=\mathrm{dim}A_f$ by the same group of people, and $\mathrm{dim}A_f \geq \frac{p-1}{2}$ (see e.g. this question), so we're done.

This whole argument works with $\mathbf{Q}(\sqrt{-1})$ replaced by any fixed imaginary quadratic field. If you demand the same dichotomy for a sequence of $A$'s of bounded dimension, I imagine it's not known.