Local maxima of Legendre polynomials When I plotted the (normalized) Legendre polynoials, I couldn't help noticing that all the local maxima lay on a really nice curve:

What is the equation of the curve (and how can we arrive to that equation)?
 A: All the local maxima do not sit on the same "nice curve". 
Under Wikipedia's normalisation with $\|P_n\|^2 = \frac{2}{2n+1}$ we can use Bonnet's recursion formula to get that 
$$ (n+1)P_{n+1}(0) = - n P_{n-1}(0) $$
Since your "normalised" $\sqrt{\frac{2}{2n+1}}\tilde{P}_n = P_n$ we get that under your normalization
$$ \frac{n+1}{n} \sqrt{\frac{2n-1}{2n+3}} |\tilde{P}_{n+1}(0)| = |\tilde{P}_{n-1}(0)| $$
The multiplicative factor is
$$ \frac{n+1}{n} \sqrt{\frac{2n-1}{2n+3}} = \sqrt{ \frac{2n^3 + 3n^2 - 1}{2n^3 + 3n^2}} < 1 $$
Hence we have that the local maximum at $x = 0$ for $n = 0\pmod 4$ is strictly increasing, and hence cannot all belong to the same curve. 

Perhaps the question you want to ask is: what is the smallest convex function defined on $(-1,1)$ that is greater than or equal to the normalised Legendre polynomials $\tilde{P}_{n}$? 
A: I claim that that curve is
$$y=\pm \frac{\sqrt{2/\pi}}{ \sqrt[4]{1-x^2}}.$$
This argument will not be rigorous, and will cite a source I haven't fully understood.
Take a look at Whittaker and Watson, A course in Modern Analysis, p. 316. They write:
$$P_n(\cos \theta) = \frac{4}{\pi} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)}  \left( \frac{\cos[(n+1/2) \theta - \pi/4]}{(2 \sin \theta)^{1/2}} + \frac{\cos[(n+3/2) \theta - 3\pi/4]}{2(2n+3) (2 \sin \theta)^{3/2}}+ \cdots \right)$$

... Shew also that the first few terms of this series give an approximate value of $P_n(\cos \theta)$ for all values of $\theta$ between $0$ and $\pi$ which are no nearly equal to either $0$ or $\pi$.

This approximation seems to be very good in practice. In the picture below, the blue curve is $P_5(x) \sqrt{\sin(\cos^{-1} x)}$ and the red curve is $512/(63 \sqrt{11} \pi)*\cos((11/2) \cos^{-1}(x) - \pi/4)$. (These are the same normalizations user79365 is using in his post.) 

I assume that, in more modern language, the authors are saying that this is an asymptotic series for $P_n(\cos \theta)$, valid on $(0, \pi)$. I have taken the liberties of applying some algebraic rearrangements, replacing their parameter $\phi$ by its definition and stopping the sum after two terms rather than the four they give. Also, they are using the normalization where $\int_{-1}^1 P_n^2 = 2/(2n+1)$, so you'd want to multiply by $\sqrt{(2n+1)/2}$.
For fixed $\theta$, that second term is bounded by $c/n$. The later terms (not displayed) die off even faster as $n$ grows. So
$$\tilde{P}_n(\cos \theta)\approx \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)}  \frac{\cos[(n+1/2) \theta - \pi/4]}{(2 \sin \theta)^{1/2}}$$
as $n \to \infty$. The $\sim$ over the $P$ indicates that I am now using your normalization. As long as $\theta/\pi$ is irrational, that $\cos$ term will swing between $1$ and $-1$, coming arbitrarily close to both ends. So
$$\lim \sup_{n \to \infty} \tilde{P}_n(\cos \theta) = \lim_{n \to \infty} \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \frac{1}{(2 \sin \theta)^{1/2}}$$
Taking $1/\sqrt{\sin \theta} =1/\sqrt[4]{1-x^2}$ out of everything, we need to compute
$$\lim_{n \to \infty} \frac{4}{\pi} \sqrt{\frac{2n+1}{2}} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} \frac{1}{\sqrt{2}}$$
This limit is similar to Wallis's product, and I think the second and third proofs on the Wikipedia page should be adaptable to evaluate it. I used the third proof, by Stirling's approximation:
$$ \sqrt{2n+1} \frac{2 \cdot 4 \cdots (2n)}{3\cdot 5 \cdots (2n+1)} = \sqrt{2n+1}  \frac{2^{2n} (n!)^2}{(2n+1)!} \approx \sqrt{2n+1}   \frac{2^{2n} (n/e)^{2n} (2 \pi n)}{((2n+1)/e)^{2n+1} \sqrt{2 \pi (2n+1)}}$$
$$=\frac{2 \pi n \sqrt{2n+1}}{(1+1/2n)^{2n} \cdot (2n+1)/e \cdot \sqrt{2 \pi (2n+1)}}=\frac{e \sqrt{2 \pi}}{(2+1/n)  (1+1/2n)^{2n}} \approx \frac{\sqrt{2 \pi}}{2}.$$
Putting back in the other constants gave the result I state above. 
Remark It's fun to note that
$$\int_{-1}^1 \left(\frac{\sqrt{2/\pi}}{\sqrt[4]{1-x^2}} \right)^2 dx = 2.$$
So $P_n^2$ is, on average, half the envelope above it.
Thanks to user79365 for providing the following image: The Legendre polynomials are in blue and the above curve is in red. It's also interesting to note the bunching up at values like $x=\pm 1/2$ and $x = 0$, when $\cos^{-1}(x)$ is a rational multiple of $\pi$. 

