$$f_X(x) = \frac12e^{-|x-\theta|}, -\infty<x<\infty$$
is a special case of the Laplace distribution given as follows:
$$f_X(x|\mu,\sigma)=\frac{1}{\sqrt{2}\sigma}e^{-\frac{\sqrt{2}|x-\mu|}{\sigma}},x\in\mathbb{R}$$
for $\sigma=\sqrt{2}$ and $\mu:=\theta$. To be more general, lets consider the Laplace distribution with parameters $(\mu,\sigma)$.
Consider the likelihood function for $N$ data samples:
$$L(\mu,\sigma;x)=\prod_{t=1}^N \frac{1}{\sqrt{2}\sigma}e^{-\frac{\sqrt{2}|x_t-\mu|}{\sigma}}=(\sqrt{2}\sigma)^{-N}e^{\frac{-\sqrt{2}}{\sigma}\sum_{t=1}^N |x_t-\mu|}$$
Take the log likelihood funtion as $l(\mu,\sigma;x)=\log(L(\mu,\sigma;x))$ and we get
$$l(\mu,\sigma;x)=-N\ln (\sqrt{2}\sigma)-\frac{\sqrt{2}}{\sigma}\sum_{t=1}^N |x_t-\mu|$$
Take the derivative with respect to the parameter $\mu$
$$\frac{\partial l}{\partial \mu}=-\frac{\sqrt{2}}{\sigma}\sum_{t=1}^N \frac{\partial|x_t-\mu|}{\partial\mu}$$
which is equal to
$$=\frac{\sqrt{2}}{\sigma}\sum_{t=1}^N\mbox{sgn}(x_t-\mu)$$
using the identity
$$\frac{\partial |x|}{\partial x}=\frac{\partial \sqrt{x^2}}{\partial x}=x(x^2)^{-1/2}=\frac{x}{|x|}=\mbox{sgn(x)}$$
To maximize the likelihood function we need to solve
$$=\frac{\sqrt{2}}{\sigma}\sum_{t=1}^N\mbox{sgn}(x_t-\mu)=0 \quad\quad (1)$$
For which we have two cases; $N$ is even or odd.
If $N$ is odd and we choose $\hat{\mu}=\mbox{median}(x_1,\ldots ,x_N)$, then there are $\frac{N-1}{2}$ cases where $x_t<\mu$ and for the other $\frac{N-1}{2}$ cases $x_t>\mu$, therefore $\hat{\mu}$ satisfies ($1$) and is the Maximum likelihood estimator for the parameter $\mu$
If $N$ is even, we can not simply choose one $x_t$ which will satisfy ($1$), however we can still minimize it through ranking the observations as $x_1\leq x_2\leq \ldots,x_N$ and then choosing either $x_{N/2}$ or $x_{(N+1)/2}$
In summary $\hat{\mu}=\mbox{median}(x_1,\ldots ,x_N)$ is the maximum likelihood estimator for any $N$
\begin{align}
& \text{For } \mu \le \min\{x_1,\ldots,x_n\} \text{ and } \alpha>0, \text{we have} \\[10pt]
L(\mu,\alpha) & = \frac 1 {\Gamma(\alpha)^n} \left( \prod_{i=1}^n (x_i-\mu) \right)^{\alpha-1} \!\!\! \exp \left( -\sum_{i=1}^n (x_i-\mu) \right), \\[10pt]
\ell(\mu,\alpha) & = \log L(\mu,\alpha) = -n\log\Gamma(\alpha) + (\alpha-1) \sum_{i=1}^n \log(x_i-\mu) - \sum_{i=1}^n (x_i-\mu).
\end{align}
You gave us $\alpha<1.$
That implies $\alpha-1<0,$ so that $\ell(\mu,\alpha)$ is an increasing function of $\mu$ until $\mu$ gets as big as $\min\{x_1,\ldots,x_n\}.$
Therefore $\widehat\mu = \min\{x_1,\ldots,x_n\}.$ If we didn't have the constraint that $\alpha<1,$ then this would be more complicated.
This value of $\widehat\mu$ does not depend on $\alpha$ as long as $\alpha$ remains in that interval. Therefore we can just plug in $\min$ for $\mu$ and then seek the value of $\alpha\in(0,1)$ that maximizes $\ell(\min,\alpha).$
Now we have
$$
\ell(\min,\alpha) = -n\log\Gamma(\alpha) + (\alpha-1)A + \big( \text{constant} \big)
$$
where "constant" means not depending on $\alpha.$
$$
\frac {\partial\ell}{\partial\alpha} = -n\frac{\Gamma'(\alpha)}{\Gamma(\alpha)} + A.
$$
Etc.
Best Answer
$\sum_{j=1}^{n}\left|\theta-x_j\right|$ is minimized by the median of the $x_j$’s. So you should consider the cases $n$ is odd or even.