Why does PCA assume Gaussian Distribution?
TL;DR

PCA does assume normal distribution of features  See p.55 SAS book1 or Rummel, 19702 or Mardia, 19793.
If you expect the PCs to be independent, then PCA might fail to live to your expectations.
Assuming that the dataset is Gaussian distributed would guarantee that the PCs are independent.

Long Answer
PCA doesn't assume the dataset to be Gaussian distributed
Most of the sources I have found (e.g. wikipedia) don't list Gaussian distribution as a requirement of PCA.
Moreover, it seems that Shlens himself doesn't believe that anymore:
I found 2 more versions of Shlens' tutorial: version 2 and version 3.02. The latter seems to be the current version (as Shlens' web page links to it), so I will refer only to version 3.02 in my answer.
In version 3.02, the paragraph you quoted was removed from the &quot;Summary of Assumptions&quot; section, so that currently, the section lists only the following assumptions:

Linearity
Large variances have important structure.
The principal components are orthogonal.

When might PCA fail to live to our (false) expectations?
In page 10 Shlens gives an example for when one might see the result of PCA as a failure, and then explains why PCA didn't really fail:

The solution to this paradox lies in the goal we selected for the
analysis. The goal of the analysis is to decorrelate the data, or
said in other terms, the goal is to remove second-order dependencies
in the data. In the data sets of Figure 6, higher order
dependencies exist between the variables. Therefore, removing
second-order dependencies is insufficient at revealing all
structure in the data.


i.e. PCs are guaranteed to be uncorrelated, so that's exactly what we should expect them to be. However, if we expected the PCs to be independent, then we would consider PCA to fail when the PCs aren't independent (e.g. the examples in figure 6).
(See this answer for another explanation of this paragraph.)
How can the assumption of Gaussian distribution help with our (false) expectations?
If we assume that the original dataset is Gaussian distributed (i.e. the features are jointly normally distributed), then by definition every linear combination of the original features is normally distributed.
Each of the PCs given by PCA is a linear combination of the original features. Thus, also every linear combination of the PCs is a linear combination of the original features, and so every linear combination of the PCs is normally distributed.
So, by definition the PCs are jointly normally distributed. PCA guarantees that the PCs are uncorrelated, and therefore they are also independent.
(Note that in the original paragraph quoted in the question, Shlens seems to claim that each of the original features should be normally distributed. However, I believe that was a mistake, and he actually meant that the original features should be jointly normally distributed (I deduced that's what he meant mainly from footnote 7 in page 10 in version 3.02). This answer explains why these conditions are not equivalent in the 2D case. Similarly, they aren't equivalent for any dimension $&gt;1$.)
Thus, under the assumption that the original dataset is Gaussian distributed, PCA guarantees that the PCs are independent.