What is $\sum_{k=1}^\infty \rm{sinc}^8(k)$ using the sine cardinal function?

Given the sine cardinal function, $$\rm{sinc}(x) = \frac{\sin x}x$$

for $x\neq0$. We have the nice evaluations,

$$\sum_{k=1}^\infty \rm{sinc}(k) = \sum_{k=1}^\infty \rm{sinc}^2(k)=-\tfrac12+\tfrac12\pi$$ $$\sum_{k=1}^\infty \rm{sinc}^3(k)=-\tfrac12+\tfrac38\pi$$ $$\sum_{k=1}^\infty \rm{sinc}^4(k)=-\tfrac12+\tfrac13\pi$$ $$\sum_{k=1}^\infty \rm{sinc}^5(k)=-\tfrac12+\tfrac{115}{384}\pi$$ $$\sum_{k=1}^\infty \rm{sinc}^6(k)=-\tfrac12+\tfrac{11}{40}\pi$$

then the not-so-nice,

$$\sum_{k=1}^\infty \rm{sinc}^7(k)=-\tfrac12+\quad\\ \tfrac{1}{46080}(129423\pi-201684\pi^2+144060\pi^3-54880\pi^4+11760\pi^5-1344\pi^6+64\pi^7)$$

However, I found this can be prettified as,

$$\sum_{k=1}^\infty \rm{sinc}^7(k)=-\frac12+\frac{7\cdot29^2\,\pi}{2^5\,6!}+\frac{\pi\big(\pi-\tfrac72\big)^6}{6!}$$

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Questions:

1. Why is the closed-form for $n=7$ far more complicated than $n<7$? (And a good lesson that "patterns" may be illusory.)
2. What is $n=8$ in terms of $\pi$? (Maybe also for $n=9$?)

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Update: Courtesy of Oliver Oloa's comment, for $n=8$, after some tweaking is,

$$\sum_{k=1}^\infty \rm{sinc}^8(k)=-\frac12+\frac{151\pi}{630}-\frac{\pi\big(\pi-\tfrac82\big)^7}{7!}$$

but $n=9$ is more complicated. See second answer below.

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