What is the CDF of the Shapiro-Wilk $W$ statistic? I am trying to understand how Shapiro-Wilk test functions. So far, I have come across the following links:
How are the values in the Shapiro Wilk weight table calculated?
which explains how to calculate the coefficients in the shapiro-wilk test. And then:
Shapiro-Wilk Test: Testing for Normality 
which explains the routine to calculate the  test-statistic and presents  a script for calculating the p-value. In my mind, I am comparing the test to a standard t-test where the test statistics has a certain CDF and it helps us to calculate the p-value. I cannot figure this out in the case of Shapiro-Wilk test. 
 A: tl;dr: There is no known distribution. i.e. it doesn't have a name. You use simulations instead.
I found this to give some insight:
Basically, Shapiro and Wilk calculated the distribution of their statistic only for $n=3$, to be a truncated $Beta(\frac{1}{2}, \frac{1}{2})$ distribution for $\frac{3}{4} \le w \le 1$ and zero elsewhere. 
For $3 < n \le 20$, the coefficients values in the statistics (expected values of ordered statistic, times inverse of co-variance matrix, normed) were calculated precisely, but there's no name for the distribution. For $20 < n \le 50$ these coefficients were only approximated, and (monte carlo) simulations were used. 
Some researchers after tried to find the asymptotic distribution, which in itself has no name.
Leslie, Stephens, and Fotopoulos showed that $n(w - \mathbb{E}(w)) \sim -\sum_{k=3}^{\infty}\frac{Z_k^2-1}{k}$ where $Z_k \sim N(0,1)$ and were i.i.d.
But in any case that too was shown to converge 

"at a painfully slow rate. ... For such reasons, the limit
  distribution seems not to be of much practical use, and for $n > 50$,
  Monte Carlo simulation seems still to be needed."

