The Alexandrov maximum principle indeed holds in your setting. The key is that the concave envelope of $u$ is in $C^{1,\,1}$, so the area formula is valid for its gradient. Assuming for simplicity that $L = \Delta$, that $\Omega = B_1$ and that $\sup_{\partial B_1} u = 0$, the way I would argue is:

Let $\Gamma$ be the concave envelope (the infimum of linear functions larger than $u$ in $B_1$ and, say, larger than $0$ on $\partial B_2$). Using that $u$ is touched from below by a paraboloid with Hessian $-\lambda^{-1}I$ at every point, one can show (see e.g. the book of Caffarelli-Cabre on fully nonlinear equations) that $\Gamma$ is touched from above by a linear function and from below by a paraboloid of opening $-\lambda^{-1}I$ at every point. In particular, $\Gamma \in C^{1,\,1}$, so we can apply the area formula to $\nabla \Gamma$. By elementary geometry $\Gamma$ is touched from above on the set $K := \{u = \Gamma\}$ by a linear function of slope $p$ for every $p \in B_{\frac{\sup_{B_1}u}{5}},$ so
$$(\sup_{B_1}u)^n \leq C(n)|\nabla \Gamma(K)|.$$
Using the area formula and that $D^2u \leq D^2\Gamma \leq 0$ at almost every point in $K$ we conclude that
$$(\sup_{B_1}u)^n \leq C(n) \int_{K} |\det D^2\Gamma(x)|\,dx \leq C(n)\int_K |\det D^2u(x)|\,dx.$$
Finally, by the AGM inequality and the equation, the last term is controlled by $C(n)\|f\|_{L^n(K)}^n$, completing the proof.