I would like to outline a proof of this famous identity, which is closely related to the question I have posted on MathOverflow. It is somewhat different from the proof of Borwein brothers.
Addendum: I can use a construction given by Mazur-Swinnerton-Dyer and Zagier to prove Ramanujan's identity for $n=37$.
Let $n=58$.
1. Following Borwein brothers, we can get
$$\frac{1}{\pi}=\sum_{m=0}^{\infty}(2\sqrt{n}v(k)m+G_0)b_mc^m(k)$$
where
$$b_m=\frac{(4m)!}{4^{4m}(m!)^4}$$
$$2v(k)=\left(1-\frac{2}{((k^{\prime})^2/(2k))^2+1}\right)$$
$$c(k)=\left(\frac{2}{2k/(k^{\prime})^2+(k^{\prime})^2/(2k)}\right)^2$$
$$G_0=\frac{\sqrt{n}}{3}\left(1-\frac{3}{2(((k^{\prime})^2/(2k))^2+1)}-\frac{1}{1+k^2}\frac{G_1}{2}\right)$$
$$G_1=\frac{nP(q^n)-P(q)}{(2K(k)/\pi)^2}$$
2. Following H. M. Weber, one has
$$\frac{2k}{(k^{\prime})^2}=\left(\frac{\sqrt{29}-5}{2}\right)^6$$
where $k=k(e^{-\pi\sqrt{58}})$
3. $$\frac{G_1}{1+k^2}=\frac{nP(q^n)-P(q)}{\eta^{2}(q^{2n})\eta^{2}(q^{4})}\frac{c(k)^{1/4}}{8\sqrt{n}}$$
Denote $$H(q)=\frac{nP(q^n)-P(q)}{\eta^{2}(q^{2n})\eta^{2}(q^{4})}$$
Then $H(q)^2$ is a weakly modular form on $\mathbb{H}/\Gamma_0(58)$
4. Denote $$[a_1,\cdots,a_n]=\prod_{\delta\mid N,\sum a_\delta=0}\eta^{a_\delta}(\delta\tau)$$
where $\eta$ is Dedekind eta function. Then $H(q)^2\cdot[-2,8,10,-16]$ is holomorphic on $\mathbb{H}/\Gamma_0(58)$ except at infinity. Then it is a linear combination of eta function product invariant under $\Gamma_0(58)$.
5. $[\cdots,\alpha_\delta,\cdots]$ is invariant under $\Gamma_0(58)$ if 1)$24\mid\sum_{\delta\mid 58}\delta a_{\delta}$; 2) $24\mid\sum_{\delta\mid 58}58a_{\delta}/\delta $ ; 3)$\prod_{\delta \mid n}\delta^{a_\delta}$ is a rational square. What's more, eta product is holomorphic at a cusp $c/d$ if
$$\frac{1}{24}\sum_{\delta\mid 58}\frac{(\mathrm{gcd}(d,\delta))^2}{\delta}a_\delta\geq 0$$
6. Expand $H(q)^2\cdot[-2,8,10,-16]$ and eta products, use matkerint function in PARI/GP to calculate the coefficients of linear combination. Note that
$$[a,b,c,d]=2^{c/2}58^{(a+b)/4}\left(\frac{\sqrt{2}}{2}\frac{\sqrt{29}+5}{2}\right)^{(a+d)/2}$$
where $q=\exp(-\pi/\sqrt{58})$. All these would lead to
$$G_0=\frac{\sqrt{58}}{3}\left(1-\frac{3}{4\times 99^2}\left(\frac{\sqrt{29}-5}{2}\right)^6-\frac{36\sqrt{2}(148 + 11 \sqrt{29})}{99\times16\sqrt{58}}\right)$$
and we are done.
Best Answer
There is a constant $C$ such that
$$\sum_{n=0}^{\infty} \frac{(\frac14)_n^3(\frac14 - k)_n}{(1)_n^3(1+k)_n} (8n+1) = C \frac{\Gamma(\frac12+k) \Gamma(1+k)}{\Gamma^2(\frac34+k)}$$
Proof: WZ-method + a Carlson's theorem (see this paper).
Then, taking $k=1/4$ we see that the only term inside the sum which is not zero, is the term for $n=0$ which is equal to $1$. This allow us to determine $C$, and we get $\, C=2 \sqrt{2}/\pi$.
Finally taking $k=0$, we obtain the value of the sum of that Ramanujan series.