I agree there is a missing $\pi$. By the integral representation for the $\zeta$ function
$$\zeta(n)-1 = \int_{0}^{+\infty}\frac{x^{n-1}}{(n-1)!}\cdot\frac{dx}{e^x(e^x-1)} \tag{1}$$
holds for any $n>1$. In particular
$$ \sum_{k\geq 1}\left(\zeta(4k)-1\right) = \int_{0}^{+\infty}\frac{dx}{e^x(e^x-1)}\sum_{k\geq 0}\frac{x^{4k+3}}{(4k+3)!}=\int_{0}^{+\infty}\frac{\sinh(x)-\sin(x)}{2e^x(e^x-1)}\,dx\tag{2} $$
where $\int_{0}^{+\infty}\frac{\sinh(x)}{2e^x(e^x-1)}\,dx=\frac{3}{8}$ and $\int_{0}^{+\infty}\frac{\sin(x)}{2e^x}\,dx=\frac{1}{4}$ are trivial and
$$\int_{0}^{+\infty}\frac{\sin(x)}{e^x-1}\,dx=\sum_{m\geq 1}\frac{1}{m^2+1}\tag{3} $$
due to the Laplace transform. The last series is well-known to be equal to $\frac{\pi\coth \pi-1}{2}$, for instance through the Poisson summation formula. Rearranging proves the claim (the second one).