Can a martingale converge in $L^1$ but not almost surely?

Yes, for a martingale convergence in $L^1$ implies almost sure convergence. Suppose $(M_n) \rightarrow M_\infty$ in $L^1$ where $(M_n)$ is a martingale. Then $\sup_n \mathbb{E}[|M_n|] < \infty$ so by Doob's martingale convergence theorem there exists $N_\infty \in L^1$ such that $(M_n) \rightarrow N_\infty$ almost surely. But $L^1$ convergence implies a.s. convergence along a subsequence so we can also find a subsequence $(M_{n_k})$ such that $(M_{n_k}) \rightarrow M_\infty$ almost surely, so we must have $N_\infty = M_\infty$ a.s.

Another, somewhat shorter proof: As $L^1$ convergence is equivalent to uniform integrability and convergence in probability, almost sure convergence follows from the convergence theorem for uniformly integrable martingales.