Is there a topology that makes every basic sequence null?

The answer is negative in every non reflexive space. If $X$ is non reflexive, there is a normalized basic sequence $(z_n)$ in X s.t. $(z_1 - z_n)_{n=2}^\infty$ and $(z_1 + z_n)_{n=2}^\infty$ are both basic sequence (necessarily semi normalized). If $x^*$ tends to zero along both of these basic sequences, then $\langle x^*, z_1\rangle =0$.

Actually, $z_1$ can be any unit vector, so $F=\{0\}$. Take a non reflexive subspace $E_1$ of $E$ that does not contain $z_1$ and let $(z_n)_{n=2}^\infty$ be a normalized type $\ell^+$ basic sequence in $E_1$.

(A basic sequence $(y_n)$ is type $\ell^+$ provided there is a constant $\delta>0$ s.t. whenever $(a_n)$ is sequence of non negative scalars, only finitely many of which are not zero, then $$ \| \sum a_ny_n\| \ge \delta \sum |a_n|. $$ Non reflexivity is equivalent to containing a normalized type $\ell^+$ basic sequence.)