Here is an elementary argument. By Weinstein's tubular neighbourhood theorem we can arrange that locally $L_1$ is the zero section of $T^* \mathbb{R}^n$, and $L_2$ is another Lagrangian intersecting $L_1$ transversely in the fiber over $0$. Trivializing $T^* \mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n$ with coordinates $(q, p)$, this means that we can write $L_2$ as the graph $\{(\xi(p), p)\}$ of the function $\xi(p) = (\xi_1(p), \ldots, \xi_n(p))$ of the fiber coordinate $p$. The condition that this graph is Lagrangian is equivalent to the requirement that the matrix of partials of $\xi$ is symmetric (this is just a version of the well-known fact that the graph of a 1-form $\eta : X \to T^*X$ is Lagrangian iff $d \eta = 0$).

By the Poincaré lemma we can therefore find a function $H(p) : \mathbb{R}^n \to \mathbb{R}$ so that $\frac{\partial H}{\partial p_i} = -\xi_i(p)$. Viewing $H$ as a function on $T^* \mathbb{R}^n$ the corresponding Hamiltonian vector field is
$$
X_H = \frac{\partial H}{\partial p_i} \frac{\partial}{\partial q_i}
= - \xi_i(p) \frac{\partial}{\partial q_i}.
$$
The time-1 flow under the Hamiltonian isotopy generated by $X_H$ thus defines a symplectomorphism which fixes $L_1$ and takes $L_2$ to the fiber of $T^* \mathbb{R}^n$ over $0$. Identifying my $T^* \mathbb{R}^n$ with your $\mathbb{C}^n$ gives the desired chart.