Yes, there is such a sequence in any non-reflexive Banach space.

It is known$^\#$ that if $Y$ is not reflexive, then $Y$ contains a basic sequence that is not boundedly complete. So if $X$ is not reflexive, then there is a semi-normalized basic sequence $(x_n^*)_{n=1}^\infty$ in $X^*$ s.t. $\sup_n \|\sum_{k=1}^n x_k^*\| < \infty$. Assume that $(x_n) \subset X$ is biorthogonal to $(x_n^*)$ (else you are done). Let $x_0^*$ be a weak$^*$ cluster point of $(\sum_{k=1}^n x_k^*)_n$. Necessarily $\langle x_0^* , x_n\rangle =1 $ for all $n\ge 1$, but $x_0^*$ is not in the normed closed span of $(x_n^*)$ because $(x_n^*)_{n=1}^\infty$ is basic, and hence $(x_n^*)_{n=0}^\infty$ is basic and not biorthogonal to any sequence in $X$.

# It is inconvenient right now for me to look for a reference for this result. It certainly is in Ivan Singer's book on bases; probably in volume one. It is almost proved in Albiac-Kalton but not stated there.