Let the random variables $X_1,X_2,\ldots,X_n$ iid $U[\theta-1\,,\theta+1]$. So the likelihood function therefore has the form:
$$L(\theta\mid X)=\prod_{i=1}^nf(X_i\mid \theta)=\frac{1}{2^n}I(X_1, \ldots , X_n \in [\theta-1\,,\theta+1])\text{ ?}$$
It is correct ?
On
It is correct, but can be further simplified by considering the order statistics of the sample. That is to say, if $$\boldsymbol x = (x_1, \ldots, x_n)$$ is the sample, then consider $$x_{(1)} = \min_i x_i, \quad x_{(n)} = \max_i x_i,$$ the first and last order statistics which are equal to the smallest and largest observations in the sample, respectively. Then the condition $$\mathbb 1(x_1, \ldots, x_n \in [\theta - 1, \theta + 1])$$ is equivalent to the simpler condition $$\mathbb 1(\theta - 1 \le x_{(1)} \le x_{(n)} \le \theta + 1),$$ because if the smallest observation is at least $\theta - 1$ and the largest is at most $\theta + 1$, then all of the other observations must also fall within the stated interval. This also shows that a sufficient statistic for $\theta$ is the ordered pair $(x_{(1)}, x_{(n)})$, meaning that in the case of $n > 2$, we can achieve data reduction by discarding all but the minimum and maximum observations.
Your likelihood function is correct.