If laplacian $\Delta_n$ is defined just as $\[\Delta_n f\](u)=\sum_v (f(v)-f(u))$, where the sum is over all neighbors v of the vertex u, then the answer is NO. Informal reason is that such $\Delta_n$ does not feel the embedding of the graph into the manifold $M$ at all.
More promising approach is to define $\Delta_n$ as $\[\Delta_n f\](u)=\sum_v c(uv)(f(v)-f(u))/m(u)$ with some edge weights $c(uv)$ and vertex weights $m(u)$ depending on the embedding of the graph into $M$. For right choice of weights the answer is positive in some particular cases: e.g., square lattice in the plane, isoradial triangulation of the plane, regular triangulation of the plane. In general there is no ``natural'' laplacian $\Delta_n$ such that convergence holds.
Here are some references, see also references therein:
[1] M. Wardetzky, S. Mathur, F. Kaberer, E. Grinspun, Discrete Laplace operators: no free lunch, Eurographics Symp. Geom. Processing, A.~Belyaev, M.~Garland (eds.), 2007.
[2] D. Chelkak and S. Smirnov, Discrete complex analysis on isoradial graphs, Adv. Math., to appear, https://arxiv.org/abs/0810.2188v2.
[3] R. Courant, K. Friedrichs, H. Lewy, Uber die partiellen Differenzengleichungen der mathematischen Physik, Math. Ann. 100, (1928), 32--74. English transl.: IBM Journal (1967), 215--234. Russian transl.: Russ. Math. Surveys 8 (1941), 125--160. https://web.stanford.edu/class/cme324/classics/courant-friedrichs-lewy.pdf
