How do I do stoichiometry for 2 limiting reactants
Your balance equation is correct except for not simplifying it. Simplified equation should be:
 $$\ce{2Al + 2NaOH + 6H2O ⟶ 2Na[Al(OH)4] + 3H2}$$

I also think you're getting a little bit confused about your equation. You probably think reacting $\ce{Al}$ and $\ce{NaOH}$ with same mole ratio (2:2) make them two limiting reagents. If that is the case, you are absolutely wrong. As M. Farooq commented elsewhere, the one reactant, which is present in the least amount in a stoichiometric sense, is the limiting reactant. Let's see your question in hand:

The amount of $\ce{NaOH}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{35 g}\; \ce{NaOH}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{40 g}\; \ce{NaOH}}  = \pu{4.38 mol}$

The amount of $\ce{Al}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{2.5 g}\; \ce{Al}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{27 g}\; \ce{Al}}  = \pu{0.463 mol}$

Since $\ce{NaOH}$ and $\ce{Al}$ react with 1:1 ratio according to the equation, $\ce{Al}$ is clearly the limiting reagent.

Hence, $\pu{0.463 mol}$ of $\ce{Al}$ reacts with $\pu{0.463 mol}$ of $\ce{NaOH}$ if you added enough water to $\pu{500 g}$ of Drano container. The leftover amount of $\ce{NaOH}$ would be: $\pu{(4.38 - 0.463) mol} = \pu{3.92 mol} = \pu{3.92 mol} \times \frac{\pu{40 g}\; \ce{NaOH}}{\pu{1 mol}} = \pu{157 g}$

Now, how much water you need? You need more than $\pu{0.463 mol}\times \frac{6}{2} = \pu{1.39 mol}$ of water, which is more than $\pu{25.0 g}$ (to be excess).

Note: Similar calculations would appied to find out amount of $\ce{H2}$ released by the reaction.