What are the side effects of adding dry ice to a swimming pool?
Solubility of $\ce{CO2}$ in water is $\pu{1.45 g/L}$ at SATP. Apparent dissociation constant for carbonic acid at the same conditions is $K_\text{a}(\ce{H2CO3}) = \pu{4.47e−7 mol L^{-1}}$. 

Now lets speculate how much dry ice could be disposed and what the pool capacity might be, guessing that the water warmed back up. It's hard to tell which pool size your apartment complex (hopefully not this one) might have, so I'm going to propose some values (marked in $\mathrm{\color{red}{red}}$) which look plausible to me. 

Assuming $m(\ce{CO2}) = \color{red}{\pu{10 kg}}$, and the pool volume of approx. $V = \color{red}{\pu{5e5 L}}$ (olympic pool is $\pu{2.5e6 L}$, I took $\frac{1}{50}$ of that), one notices that in order to completely dissolve $\pu{10 kg}$ of $\ce{CO2}$, $\pu{6.9e3 L}$ of water is required (based on solubility data), which means that we have more than enough water in the pool to completely dissolve all $\ce{CO2}$. 

For the weak acid 

$$[\ce{H+}] \approx \sqrt{K_\text{a} C} = \sqrt{\frac{K_\text{a} m}{M V}} = \sqrt{\frac{\pu{4.47e−7 mol L^{-1}} \cdot \pu{e4 g}}{\pu{44 g mol ^{-1} \cdot \pu{5e5 L} }}} = \pu{1.43e-5 mol L^{-1}}$$

so the pool water will indeed be acidified to the level of

$$\mathrm{pH} = -\log[\ce{H+}] = 4.85,$$

assuming the values I've willfully established, and that all carbon dioxide completely dissolved in water. You can put your values into the equation and argue with your landlord further.