Analysis

This problem examines the classical probability model and the determination of function roots. The key is to analyze the relationship between $m$ and $n$ based on the method of determining function roots. If there is a root, then $f(1)f(2) \leqslant 0$. Solving this gives $-2m-8 \leqslant n \leqslant -1-m$. Discussing by cases and using the classical probability formula to calculate the probability.

Solution

Since $f(x) = x^3 + mx + n$, we have $f'(x) = 3x^2 + m$.

Given $m \in \{1,2,3,4\}$, we have $f'(x) > 0$,

Thus, the function $f(x)$ is increasing in the interval $[1,2]$.

If there is a root, then $f(1)f(2) \leqslant 0$, solving this gives $-2m-8 \leqslant n \leqslant -1-m$.

Therefore, the cases that allow the function to have a root in the interval $[1,2]$ are:

- For $m=1$, $-10 \leqslant n \leqslant -2$, thus $n=-8$, $n=-4$, $n=-2$, totaling $3$ cases;

- For $m=2$, $-12 \leqslant n \leqslant -3$, thus $n=-12$, $n=-8$, $n=-4$, totaling $3$ cases;

- For $m=3$, $-14 \leqslant n \leqslant -4$, thus $n=-12$, $n=-8$, $n=-4$, totaling $3$ cases;

- For $m=4$, $-16 \leqslant n \leqslant -5$, thus $n=-12$, $n=-8$, totaling $2$ cases.

Thus, there are a total of $3+3+3+2=11$ cases where the function has a root.

Since there are $4 \times 4 = 16$ possible functions,

According to the classical probability model, the probability of having a root is $\dfrac{11}{16}$.

Therefore, the correct choice is $\boxed{C}$.