$(2019_{11}) !$

In base $10$ , $2019_{11}$ is $2 \cdot 11^3 + 0 \cdot 11^2 + 1 \cdot 11^1 + 9 \cdot 11^0 = 2682_{10}$ , so $(2019_{11})! = (2682_{10})!$ .

The number of trailing zeros a number has in base $11$ is the same as the number of factors of $11$ it has. Since $2682! = 2682 \cdot 2681 \cdot 2680 \dots 3 \cdot 2 \cdot 1$ , $\text{int}(\frac{2682}{11}) = 243$ of these numbers are multiples of $11$ , $\text{int}(\frac{2682}{11^2}) = 22$ of these numbers are multiples of $11^2$ , and $\text{int}(\frac{2682}{11^3}) = 2$ of these numbers are multiples of $11^3$ , so $2682!$ has at most $243 + 22 + 2 = 267$ factors of $11$ , which means there are $\boxed{267}$ trailing zeros in $(2019_{11})!$