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You must be more careful in deducing the likelihood function, otherwise your thinking seems correct. The density function for one $X_i$ is $$ f(x_i)=\frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} $$ So the likelihood function can be written as \begin{align*} \mathcal{L}(\theta)&=\prod_{i=1}^n \frac1{2\theta}\cdot I\{ -\theta<X_i < \theta \} \\ &=2^{-n}\theta^{-n} \cdot \prod_i I\{-\theta<X_i < \theta \} \\ &= 2^{-n}\theta^{-n} \cdot I\{-\theta < \min_i x_i \le \max_i x_i <\theta \} \\ &= 2^{-n}\theta^{-n} \cdot I\{0\le \max[|\min_i x_i|, |\max_i x_i|]\} < \theta \end{align*} and this is zero if $\theta$ is too small, that is, lesser than $\max[|\min_i x_i|, |\max_i x_i|]$. At that point it becomes positive, and then decreasing from there. That gives the MLE as $$ \hat{\theta}_\text{MLE}=\max[|\min_i x_i|, |\max_i x_i|] $$ Another way to the solution is by noting that $|X_i| \sim \mathcal{U}(0,\theta)$ and use the solution for that case (which leads to an equivalent likelihood function), se for instance here