The summation is the key

Consider the Dirichlet beta function $\beta(x) \; = \; \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^x}$ Then $\beta'(x) \; = \; \sum_{n=0}^\infty \frac{(-1)^{n+1}\ln(2n+1)}{(2n+1)^x}$ and hence $\beta'(1) \; = \; \sum_{n=0}^\infty \frac{(-1)^{n+1}\ln(2n+1)}{2n+1}$ and so the infinite product we want is $\begin{aligned} e^{\beta'(1)} & = \mathrm{exp}\Big[\tfrac14\pi\Big(2\ln2 + 3\ln\pi - 4\ln\Gamma(\tfrac14) + \gamma\Big)\Big] \; = \; \mathrm{exp}\Big[\tfrac14\pi\Big[\ln\Big(\frac{4\pi^3}{\Gamma(\frac14)^4}\Big) + \gamma\Big]\Big] \; = \; \mathrm{exp}\Big[\tfrac14\pi\Big[\ln\Big(\frac{\Gamma(\frac34)^4}{\pi}\Big) + \gamma\Big]\Big] \end{aligned}$ making the answer $4+3+4+4=\boxed{15}$ .

Let us concatenate the product as follows, naming the product $P$ .

$P = \displaystyle \prod\limits_{k=1}^{\infty} (2k+1)^{(-1)^{k+1} / (2k+1)}$

If we'll take the logarithm of both sides, we'll get

$\ln P = \displaystyle \sum\limits_{k=1}^{\infty} (-1)^{k+1} \frac{ \ln (2k+1)}{ 2k+1}$

The Malmsten-Kummer series expansion for $\gamma$ is as follows:

$\gamma = \ln \pi - 4 \ln \big( \Gamma (\frac{3}{4}) \big) + \frac{4}{\pi} \cdot \displaystyle \sum\limits_{k=1}^{\infty} (-1)^{k+1} \frac{ \ln (2k+1)}{ 2k+1}$

From here we can now see that

$\displaystyle \sum\limits_{k=1}^{\infty} (-1)^{k+1} \frac{ \ln (2k+1)}{ 2k+1} = \frac{\pi}{4} \bigg[ \ln \big( \frac{[\Gamma(\frac{3}{4})]^4}{\pi} \big) + \gamma \bigg]$

Giving us $4+3+4+4 = 15$ .