This is not a a well posed question because the rules of what expressions are allowed have not been specified. If, for example, the domain is expressions of the form $\frac{\sqrt{a}+\sqrt{b}}{c}$ with $a,b,c$ non-negative integers then one could call an expression a best approximation (to $\pi \approx 3.1415926536$) if it has smaller error than any $\frac{\sqrt{a'}+\sqrt{b'}}{c'}$ with $c' \le c.$ Then one could ask how to find the best approximations, is your expression one of these and , if so, is it a better approximation than one might expect for a bound $c \le n?$ For rational numbers $\frac{p}{q}$ it is known how to find the best (rational) approximations and that one can expect from them that the approximation will have error roughly $\frac{1}{q^2}.$ A particular best rational approximation mentioned by Pietro is $\frac{355}{113}.$ See my comment for why it could be considered surprisingly good. I previously guessed that for your problem (as I have framed it) one might expect $\frac{1}{q^6}$ error but now I think that is wrong. See below. By this measure, the expression given ,$$\frac{29\sqrt{6}+\sqrt{870}}{32}=\frac{\sqrt{5046}+\sqrt{870}}{32}\approx 3.1415926546$$ is fairly good but $$\frac{3\sqrt{41}+\sqrt{149}}{10}=\frac{\sqrt{123}+\sqrt{149}}{10} \approx 3.1415928328$$ is better (relative to the denominator) as are $\frac{10+\sqrt{229}}{8}$ and $\frac{1+\sqrt{71}}{3}.$ None of these impress me as much as $355/133$ though.
later thoughts This is fun as a puzzle but not much more. $\pi$ is an exceptional number and has particular approximation expressions, but these are not among them. Best rational approximation and continued fractions are quite special. The approximations are easy to find, can actually be useful, and certain accuracy can be certain. They have even been suggested as a possible alternative to floating point for use in computer computations with reals. The arithmetic and geometry are beautiful and the mathematical connections are deep. It is not a coincidence that the first few approximations to $\pi$ are $\frac31,\frac{22}{7}=\frac{1+7\cdot 3}{0+7\cdot 1},\frac{333}{106}=\frac{3+22\cdot 15}{1+7\cdot 15}$ and $\frac{355}{113}=\frac{22+333}{7+106}$. The accuracy of an approximation depends only on the fractional part (so it as easy or hard to get $\pi$ with a denominator under $n$ as to get $100+\pi$ ). None of these things seem to be true for $\frac{\sqrt{a}+\sqrt{b}}{c}$ nor for roots of degree 4 polynomials.
That said, I now think that to approximate a positive target real $T$ with denominator exactly $c$ one can expect an error of order $\frac{1}{c^4T^3}$ This because the number of expressions $\sqrt{a}+\sqrt{b}$ in an interval $(x-1/2,x+1/2)$ is almost exactly $\frac{x^3}{3}$ so we would expect to be able to approximate $cT$ by $\sqrt{a}+\sqrt{b}$ with error of order $\frac{1}{c^3T^3}$ and hence $T$ by $\frac{ \sqrt{a}+\sqrt{b}}{c}$ with accuracy as given. So I propose defining the virtue of an approximation $\frac{ \sqrt{a}+\sqrt{b}}{c}$ to $T=\pi$ to be $-log_{c\pi}|\pi-\frac{ \sqrt{a}+\sqrt{b}}{c}|$ and expect it to be about $3$. I can report that for $3 \le c \le 200$ the $198$ virtue values of the best approximations (one for each $c$) are best fitted by the line $3.0339-0.000014c$ so that certainly seems satisfyingly flat. I find (in accordance with more extensive reports by others here) that the approximations which beat any previous one (with regard to absolute error) are for these triples $[c,a,b]=$ $\small [3, 1, 71], [4, 38, 41], [5, 45, 81], [6, 2, 304], [6, 5, 276], [7, 18, 315], [8, 100, 229], [10, 149, 369], [14, 181, 932]$
$\small [21, 469, 1964], [24, 120, 4153], [27, 937, 2939], [28, 1724, 2157], [31, 576, 5386], [32, 870, 5046], [59, 2027, 19693]$
$ \small [69, 930, 34698] [80, 697, 50592], [91, 9774, 34977], [98, 2377, 67144], [120, 2010, 110329]$
$\small [132, 1311, 143249], [142, 14503, 106066], [152, 36835, 81566], [181, 67364, 95532]$
For these 24 best approximations the virtues get up to 3.83837,3.80356,3.734 at c=10,8,32 respectively. However these are relatively early so it is hard to say what to expect.
I believe the comments of joro and Carlo Benakker right under your question is to the point. Since the zeroes close to $s$ will be the ones that contributes in the sum, the zeroes close to s must be computed first and in order to do that several values of the zeta-function must certainly be calculated and the time for each such computation (and zero) will certainly not be less than computing any other particular value of $\zeta(s)$ or $\zeta'(s)$.
As for calculating a particular value $\zeta(s)$ the recent method of Ghaith Hiary http://arxiv.org/abs/0711.5005 "Fast methods to compute the Riemann zeta function" (published in Annals of Mathematics 2011) contains the fastest method known. He shows a method that calculates the value of $\zeta(1/2+it)$ with an error term less than $|t|^{-N}$ in time $O_{\varepsilon,N}(t^{4/13+\varepsilon})$. The method of Odlyzko and Schönhage "Fast algorithms for multiple evaluations of the Riemann zeta function", doi:10.2307/2000939 is faster if sufficiently many values of $t$ needs to be calculated, i.e. it can calculate $\sqrt T$ values in the range $[T,T+\sqrt T]$ in time $O_\varepsilon(T^{1/2+\varepsilon})$.
Now, In Hiary's paper it might look like he only considers the critical line. However in fact his method works for any value in the critical strip. Indeed in part of his argument he considers just the line Re$(s)=0$, but then he writes "It is clear that the restriction $\sigma=0$ is not important and a similar conclusions can be drawn for other values of $\sigma$" (page 6, second to last paragraph in Hiary's paper). It is true that his main interest is the critical line, but the algorithm holds for any $s$ in any vertical strip such as the critical strip.
Now to calculate $\zeta'(s)$ up to error of order $t^{-N}$ it is sufficient to calculate $\zeta(s+ih)$ and $\zeta(s-ih)$ with error less than $t^{-3N/2-1/4}$ for $h=t^{-N/2-1/4}$ and consider the difference quotient $(\zeta(s+ih)-\zeta(s-ih))/h$ by also using that $\zeta'''(s) =O(\sqrt t)$ in the critical strip and some version of the mean value theorem inequality.
It is simple to use finite difference quotients to calculate $\zeta^{(k)}(s)$ up to any desired degree of accuracy. Thus we can certainly use this method to calculate for example
$$
\frac d {ds} \frac{\zeta'(s)}{\zeta(s)}= \frac{\zeta''(s)\zeta(s)-\zeta'(s)^2}{\zeta(s)^2}
$$
by this method. This method should not be computionally wasteful.
Best Answer
This has been answered in the comments by Lucia. The identity $$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}}}$$ is false. By subtracting $1$ from both sides and squaring, we have that $$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=1-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}},$$ which by the identity implies that $$(1-P(s))^{2}+1-\frac{2}{\zeta(s)}=P(2s).$$ This equation, which appears as theorem $1$ in the paper, is not true.