This infinite product could be computed (more or less) directly.
At first, let's compute infinite sum instead of infinite product:
$$P=\prod_{n=1}^\infty \left(1+\frac2n\right)^{(-1)^{n+1}\,\,n}$$
$$S = \ln P = \sum_{n=1}^\infty (-1)^{n+1}n \ln\left(1+\frac2n\right) = \sum_{n=1}^\infty (-1)^{n+1}a_n$$
At first we need some regularization:
$$a_n=2-\frac2n+\frac{8}{3n^2}+\mathcal O(\frac{1}{n^3}) \nrightarrow 0$$
Nevertheless we could only consider partial sums of the $S$ with odd (or even) $n$; in other words, we'll assume that $\sum_{n=1}^\infty (-1)^{n+1} = 0$. (I know, I know, this is a not rigorous, but let me advance).
Ok, let's convert $a_n$ to power series:
$$a_n = \sum_{k=0}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k}=2+\sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k},$$
so we need to compute
$$S = \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1}\frac{1}{n^k}$$
(look, we «cancel» 2 in $a_n$'s expansion). Let's change summation order:
$$S = \sum_{k=1}^\infty (-1)^k\frac{2^{k+1}}{k+1} \left(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^k}\right).$$
Series in the parentheses is alternating zeta-function:
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^k} = \left(1-\frac{1}{2^{k-1}}\right)\zeta(k),$$
where $\zeta(k)$ is the Riemann zeta function ($k>1$). At $k=1$ we have
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\ln2.$$
So,
$$S = -2\ln2 + \sum_{k=2}^\infty (-1)^k\frac{2^{k+1}}{k+1}\left(1-\frac{1}{2^{k-1}}\right)\zeta(k)\\=-2\ln2+\sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\zeta(k).$$
This series seems hopeless, but zeta function involves in many other series. We'll use this:
$$\Gamma(s)\zeta(s)=\int\limits_0^\infty \!\frac{x^{s-1}}{e^x-1}dx.$$
Hence
$$S + 2\ln2 = \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{1}{\Gamma(k)}\int\limits_0^\infty \!\frac{x^{k-1}}{e^x-1}dx \\=
\sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{1}{(k-1)!}\int\limits_0^\infty \!\frac{x^{k-1}}{e^x-1}dx,$$
or
$$T = S + 2\ln2 = \int\limits_0^\infty \!\frac{dx}{e^x-1} \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{x^{k-1}}{(k-1)!}$$
The last series could be computed by standard methods via series for exponent and differentiation w.r.t. $x$:
$$\sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{k+1}\frac{x^{k-1}}{(k-1)!} = \sum_{k=2}^\infty 4(-1)^k\frac{2^{k-1}-1}{(k+1)!}kx^{k-1}\\=
\frac{3}{x^2}+\frac{1}{x^2}e^{-2x}+\frac{2}{x}e^{-2x}-\frac{4}{x^2}e^{-x}-\frac{4}{x}e^{-x}$$
Let's collect all these pieces together:
$$T = \int\limits_0^\infty \frac{e^{-2x}dx}{x^2(e^x-1)}(3e^{2x}-4(1+x)e^x+2x+1).$$
We have here terms with powers of $\mathrm{exp}$ and $x$. For me simplest and straightforward method for calculation of this integral is expansion of denominator:
$$\frac{1}{e^x-1}=\frac{e^{-x}}{1-e^{-x}}=e^{-x}\sum_{m=0}^\infty e^{-mx} = \sum_{m=1}^\infty e^{-mx}.$$
So we have
$$T = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-2x}\sum_{m=1}^\infty e^{-mx},$$
or
$$T = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} \sum_{m=3}^\infty e^{-mx}.$$
Let's change summation and integration:
$$T = \sum_{m=3}^\infty \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-mx}.$$
Now we need to compute
$$I_m = \int\limits_0^\infty dx\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} e^{-mx}.$$
Maple says
$$I_m = m\ln(m) - 4m\ln(m - 1) + 3m\ln(m - 2) - 2\ln(m) + 8\ln(m - 1) - 6\ln(m - 2) + 2.$$
We could believe in it or could to get it:) At first,
$$\frac{3e^{2x}-4(1+x)e^x+2x+1}{x^2} = \sum_{k=0}^\infty \frac{12\cdot 2^k-4k-12}{(k+2)!} x^k,$$
and
$$\int\limits_0^\infty \frac{12\cdot 2^k-4k-12}{(k+2)!} x^k e^{-mx} dx = \frac{12\cdot 2^k-4k-12}{(k+1)(k+2)} \frac{1}{m^{k+1}}$$
(we used Euler's formula for $\Gamma(x)$). So,
$$I_m = \sum_{k=1}^\infty \frac{12\cdot 2^k-4k-12}{(k+1)(k+2)} \frac{1}{m^{k+1}}$$
(term at $k=0$ vanishes). Now we can diff. w.r.t. $m$:
$$\frac{d^2 I_m}{dm^2}= \sum_{k=1}^\infty (12\cdot 2^k-4k-12) \frac{1}{m^{k+3}} = \frac{1}{m^3} \sum_{k=1}^\infty \frac{12\cdot 2^k}{m^k}-\frac{4k}{m^k}-\frac{12}{m^k}, $$
or
$$\frac{d^2 I_m}{dm^2}= \frac{1}{m^3} \left(\frac{24}{m - 2} - \frac{4m}{(m - 1)^2} - \frac{12}{m - 1}\right)$$
(while $m>2$). Integrate twice and we have indeed
$$I_m = (3m - 6)\ln(m - 2) + (-4m + 8)\ln(m - 1) + 2 + (m - 2)\ln(m)\\= m\ln(m) - 4m\ln(m - 1) + 3m\ln(m - 2) - 2\ln(m) + 8\ln(m - 1) - 6\ln(m - 2) + 2.$$
We start summation from $m=3$; let $m=2+p$:
$$I_{p+2} = p\ln(2 + p) - 4p\ln(1 + p) + 3p\ln(p) + 2$$
Now we have
$$T = \sum_{p=1}^\infty p\ln(2 + p) - 4p\ln(1 + p) + 3p\ln(p) + 2 = \sum_{p=1}^\infty t_p.$$
We could find approximation for this sum by some methods, but I prefer more detailed calculation. At first, let's rearrange $t_p$:
$$t_p = [(p + 2)\ln(p + 2) - 4(p + 1)\ln(p + 1) + 3p\ln(p)] \\ +[2-2\ln(p + 2) + 4\ln(p + 1) ].$$
First sum could be found by kind of telescoping summation:
$$T^1_N = \sum_{p=1}^N (p + 2)\ln(p + 2) - 4(p + 1)\ln(p + 1) + 3p\ln(p) \\= -3(N + 1)\ln(N + 1) + (N + 2)\ln(N + 2) - 2\ln(2)$$
(put $f(x)=x\ln x$ and look at partial sums). Analogously for second term we have
$$T^2_N = \sum_{p=1}^N 2-2\ln(p + 2) + 4\ln(p + 1) \\= 2\sum_{p=1}^N \ln p + 2\ln(N + 1) - 2\ln(N + 2) + 2\ln(2) + 2N$$
Finally we have
$$T_N = 2\sum_{p=1}^N \ln p - (3N + 1)\ln(N + 1) + N\ln(N + 2) + 2N.$$
Asymptotic expansion for the first series is well-known and could be found from Euler—Maclaurin:
$$\sum_{p=1}^N \ln p = N(\ln(N) - 1) + \frac12\ln(2\pi) + \frac12\ln(N) + \frac{1}{12N}+\mathcal O \left(\frac{1}{N^2}\right),$$
and finally (really finally :)
$$T_N = \ln(2\pi)-1 - \frac{4}{3N}+\mathcal O \left(\frac{1}{N^2}\right).$$
So,
$$T = \ln(2\pi)-1,$$
$$S = T - 2\ln2 = \ln\frac{\pi}{2}-1 = \ln\frac{\pi}{2e},$$
and
$$P = \frac{\pi}{2e}.$$
