Trying to answer this question, I face the problem of approximating
$$a_n=\Big|E_{2 n-1}(1)-E_{2 n-1}(0)\Big|$$
From definition $$E_{2 n-1}(1)-E_{2 n-1}(0)=4^{1-n}\, \Gamma(2n)\,\sum_{k=0}^n \frac{E_{2 k}}{\Gamma (2 k+1) \Gamma (2 n-2 k)}$$
Empirically, it seems that $$\log\left(\Bigg|\sum_{k=0}^n \frac{E_{2 k}}{\Gamma (2 k+1) \Gamma (2 n-2 k)} \Bigg|\right) \sim \log(2) + 2 \log \left(\frac{2}{\pi }\right)\,n=\log \left(2^{2 n+1} \pi ^{-2 n}\right) \tag 1$$
For a check, a quick and dirty linear regression (for $n=1$ to $n=1000$) gives (with $R^2 =0.99999999984$ !)
$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & +0.6940519 & 4.21\times 10^{-4} & \{+0.6932257,+0.6948781\} \\ \beta & -0.9031668 & 7.29\times 10^{-7} & \{-0.9031682,-0.9031653\} \\ \end{array}$$
Comparing $$a_n=\Big|E_{2 n-1}(1)-E_{2 n-1}(0)\Big|\qquad \text{and} \qquad 8 \pi ^{-2 n} \Gamma (2 n)$$ the relative error is lower than $1.7 \times 10^{-3}$% as soon as $n \geq 5$ and lower than $2.9 \times 10^{-8}$% as soon as $n \geq 10$.
Any idea or suggestion would be more than welcome.
