Since $f(x)$ is monotonically decreasing on $[0,+\infty)$,
it means $f(x)$ is monotonically increasing on $(-\infty,0)$,
Based on the inequality $f(x-1) \geqslant f(1)$,
we get $|x-1| \leqslant 1$, solving this yields: $0 \leqslant x \leqslant 2$,
Therefore, the probability $p$ that satisfies the condition is $p= \dfrac{2}{8}= \dfrac{1}{4}$,
So, the answer is: $\boxed{\dfrac{1}{4}}$.
This problem is solved by utilizing the monotonicity and evenness of the function to derive an inequality about $x$, and then solving it using the definition of geometric probability.
This question tests the method of calculating probabilities in geometric models; the probability in geometric models is determined by the ratio of lengths, areas, or volumes, and is considered a medium-difficulty question.