It seems to me like this would be calculating the odds of a coin toss that has already happened and we know the results of,
I think it's simpler than that. Nate Silver has a model for the polling error term, and he fit its parameters (in this case, one parameter, the standard deviation) based on one of its summaries (the average of past polling errors), much like the maximum likelihood estimation method for fitting a distribution parameters given a sample taken from it.
Say that the polling error $\epsilon$ is normally distributed with mean $\mu = 0$ and unknown standard deviation $\sigma$, i.e.
\begin{equation}
\epsilon \sim \mathcal{N}(0, \sigma ^2)
\end{equation}
Now $\epsilon$ can be negative or positive, but we would like to compute the average of all polling errors' magnitudes. We can describe this new random variable $M$ representing the magnitude of the polling error as
\begin{equation}
M = \left| \epsilon \right|
\end{equation}
We don't know how it is distributed, but we can describe its cumulative density function. The CDF of $M$ is such that
\begin{align}
F_M(x) &= P(M<x) \\
&= P(\left| \epsilon \right| < x)
\end{align}
Let's expand the norm above using two expressions:
\begin{align}
F_M(x) &= P(-x < \epsilon < x) \\
&= P(\epsilon < x) - P(\epsilon < -x) \\
&= F_N(x, 0, \sigma) - F_N(-x, 0, \sigma)
\end{align}
where $F_N(x, \mu, \sigma)$ is the CDF for the normal distribution with parameters $\mu$ and $\sigma$. We see above that $F_M(x)$ depends on the standard deviation of the polling error, so we'll call it $F_M(x, \sigma)$ from now on.
Given $M$'s CDF, the probability density function (PDF) of $M$, $f_M(x, \sigma)$, can be calculated by:
\begin{equation}
f_M(x, \sigma) = \dfrac{d}{d\,x} F_M(x, \sigma)
\end{equation}
Run it through the calculus machine and we get an expression for $f_M(x, \sigma)$:
\begin{equation}
f_M(x, \sigma) = \sqrt{\dfrac{2}{\pi\, \sigma}} \cdot \exp \left(-\dfrac{x^2}{2\, \sigma ^2} \right) \qquad \qquad x \geqslant 0
\end{equation}
The average of the polling errors (which is what we went through all this for) is the mean of $M$; as such, it can be calculated using $f_M(x, \sigma)$:
\begin{equation}
\dfrac{\mu_M}{\sigma} = \int _{0} ^{+\infty} x\, f_M(x, \sigma)\, \mathrm{d} x
\end{equation}
The whole point is to estimate the polling error distribution parameter (its standard deviation) by using a summary of its sample (the average of polling errors). This last expression gives us just that, a relationship between the standard deviation of the polling error and the mean of its magnitude. Plugging in the numbers from Nate Silver's article and solving for $\sigma$, we get, for $\mu _M = 6\, p.p.$, $\sigma = 7.52\, p.p.$.
The true margin of error $\Delta$, as defined by Silver, is the half-amplitude of the $95\%$ confidence interval on the polling error:
\begin{align}
P(-\Delta < \epsilon < \Delta) &= 95\% \\
P(\epsilon > \Delta) = P(\epsilon < -\Delta) &= 2.5\% \\
1 - F_N(\Delta, 0, \sigma) &= 2.5\%
\end{align}
Doing the math,
\begin{equation}
\Delta \approx 1.96\, \sigma
\end{equation}
Thus we then get a true margin of error of plus or minus $14.7\, p.p.$, as stated in the article.
I chose to represent the polling error as a normal random variable (as is common when doing OLS regression), but there are other equally good choices, which would yield different results.
