Although you are also asking about the estimator $\hat{\lambda}$, I am going to note some things about $\hat{\theta}$. In this particular case it is quite easy to obtain the exact distribution of this estimator. Since you have a series of shifted exponential random variables, you can define the values $Y_i = X_i - \theta$ and you then have the associated series $Y_1,Y_3,Y_3 ... \sim \text{IID Exp}(\lambda)$. This gives the exact distribution:
$$\hat{\theta} = X_{(1)} = \theta+ Y_{(1)} \sim \theta + \text{Exp}(n \lambda).$$
Note that this gives the pivotal quantity $n(\hat{\theta} - \theta) \sim \text{Exp}(\lambda)$. You can prove that $\hat{\theta}$ is a consistent estimator by computing the probability of a deviation larger than a specified level. For all $\varepsilon >0$ we have:
$$\begin{aligned}
\mathbb{P}(|\hat{\theta} - \theta| < \varepsilon)
= \mathbb{P}(\hat{\theta} - \theta< \varepsilon)
= \exp(-n \lambda \varepsilon). \\[6pt]
\end{aligned}$$
We therefore get the limiting result:
$$\begin{aligned}
\lim_{n \rightarrow \infty} \mathbb{P}(|\hat{\theta} - \theta| < \varepsilon)
= \lim_{n \rightarrow \infty} \exp(-n \lambda \varepsilon) = 0, \\[6pt]
\end{aligned}$$
which is the required condition for weak consistency (i.e., convergence in probability of the estimator to the parameter it is estimating).