This is more of an extended comment than an answer, but I figure that since the OP is interested in the computational aspect ("...I wrote a little script to calculate it..."), this might be of some interest. The following is adapted from Ole Østerby's unpublished manuscript.
As already noted, the OP's limit is equivalent to saying that
$\lim_{h\to 0}\frac{\sin\,\pi h}{h}=\pi$
where we identify $h$ with $2^{-n}$. That is, to borrow notation from David's answer (and taking $r=1$):
$b_n=2^n \sin\left(\frac{\pi}{2^n}\right)$
If we expand $\dfrac{\sin\,\pi h}{h}$ as a series, like so:
$\frac{\sin\,\pi h}{h}=\pi-\frac{\pi^3 h^2}{6}+\frac{\pi^5 h^4}{120}+\cdots$
we see that only even powers of $h$ occur in the expansion (as expected, since the function in question is even).
If we halve $h$ (equivalently, increase $n$), we have a slightly more accurate approximation of $\pi$. The upshot is that one can take an appropriate linear combination of $\dfrac{\sin\,\pi h}{h}$ and $\dfrac{\sin(\pi h/2)}{h/2}$ to yield an even better approximation to $\pi$:
$\frac13\left(\frac{4\sin(\pi h/2)}{h/2}-\frac{\sin\,\pi h}{h}\right)=\pi -\frac{\pi^5 h^4}{480}+\frac{\pi^7 h^6}{16128}+\cdots$
This game can be repeatedly played, by taking successive linear combinations of values corresponding to $h/2$, $h/4$, $h/8$... The method is known as Richardson extrapolation.
(I'll note that I have already brought up Richardson extrapolation in a number of my previous answers, like this one or this one.)
More explicitly, taking $T_n^{(0)}=b_n$, and performing the recursion
$T_j^{(n)}=T_{j}^{(n-1)}+\frac{T_{j}^{(n-1)}-T_{j-1}^{(n-1)}}{2^n-1}$
the "diagonal" sequence $T_n^{(n)}$ is a sequence that converges faster to $\pi$ than the sequence $b_n$. Christiaan Huygens used this approach (way before even Richardson considered his extrapolation method) to refine the Archimedean estimates from circumscribing and inscribing polygons.
Sundry Mathematica code:
Table[2^(n - 1)*Sqrt[2 - Nest[Sqrt[2 + #1] & , 0, n - 2]] == FunctionExpand[2^n*Sin[Pi/2^n]], {n, 2, 15}] {True, True, True, True, True, True, True, True, True, True, True, True, True, True}
This verifies the equivalence of the iterated square root and sine expression.
Here is an implementation of the application of Richardson extrapolation to the computation of $\pi$:
huygensPi[n_Integer, prec_: MachinePrecision] := Module[{f = 1, m, res, s, ta}, res = {ta[1] = s = N[2, prec]}; Do[ If[k > 1, s = s/(2 + Sqrt[4 - s])]; f *= 2; ta[k + 1] = f Sqrt[s]; m = 1; Do[m *= 2; ta[j] = ta[j + 1] + (ta[j + 1] - ta[j])/(m - 1);, {j, k, 1, -1}]; res = {res, ta[1]};, {k, n - 1}]; Flatten[res]]
Note that I used a stabilized version of the recursion for generating the $b_n$ (f Sqrt[s]
in the code) to minimize errors from subtractive cancellation.
Here's a sample run, where I generate 10 successive approximations to 25 significant digits:
huygensPi[10, 25] - Pi {2.000000000000000000000000, 3.656854249492380195206755, 3.173725640962868268333725, 3.13944246625722809089242, 3.14157581875151427853903, 3.14159291451874033422144, 3.14159265388327967181647, 3.14159265358866759077617, 3.14159265358979303435864, 3.14159265358979323865872}
where the $10$-th approximant is good to 19 digits. By comparison, $b_{10}=2^{10}\sin\left(\dfrac{\pi}{2^{10}}\right)= 3.1415877\dots$ is only good to five digits.