Is Griffiths simply wrong here? (Electrostatic Boundary Conditions) 
In the above illustration, shouldn't $E_{above}$ and $E_{below}$ be in opposite directions? If not, how did Griffiths end up the following equation? From the above directions, shouldn't the flux add up?
 A: Griffiths in correct.  The flux through the top portion of the box is not just $E_{\small{top}}^\top$ but actually $\vec E\cdot (A\vec n) $ where $\hat n$ is perpendicular to the surface and points out: in your diagram, $\hat n$ would be along $+{\hat z}$ for the top portion of the flux calculation.
For the bottom of the box the $\hat n$ vector points along $-\hat z$ so the flux through through those sides of the box work out to 
$$
(E_{\small{top}}^{\top} - E_{\small{bottom}}^\top)A \tag{1}
$$
when $\vec E_{\small{bottom}}$ points along $+\hat z$: the minus sign in (1) comes from the $-\hat z$ for the direction of the surface element $d\vec A$ at the bottom of the box.
Thus, ignoring the thin sides of the box because the flux through those is arbitrarily small, you have 
$$
(E_{\small{top}}^{\top} - E_{\small{bottom}}^\top)A=\frac{1}{\epsilon_0} \sigma A
$$
and the area cancels out.
A: The illustration and the equation may be confusing. It corresponds to a situation with an external field pointing upward that is larger than that due to $\sigma$. However the two arrows are of the same length, so if this were intended then $\sigma=0$. If there is no such field then $E_{above}$ and $E_{below}$ point in opposite directions and their magnitudes should be added, not subtracted.
A: Two ways of seeing that it’s right:


*

*Consider the case of no charge. Then nothing interesting is happening at the sheet, so the fields should be equal: both sides of the equation are zero. 

*Gaus’s law: the sum of the fields going away, which is the outward flux, is given by the charge. Since $E_{below}$ is defined as towards the charge, it enters that calculation with a minus sign. 


He’s picked a sign convention where the field upward is positive everywhere. That means that $E_{below}$ is defined such that a positive value means "the E vector points up" and a negative value means "the E vector points down".  Which way the E vector points is given by the physics:


*

*if there's positive charge on the surface and no external field, the E field below it will point down, hence $E_{below}$ will have a negative value.

*If there's a large upward going external field, then in that case the E field points upward everywhere, and $E_{below}$ will be positive.


To put it another way, the length of the $E_{below}$ arrow in the picture isn't showing you the absolute magnitude; that has to come from somewhere else and is written as "$E_{below}$", a number.  And it's not even showing you the direction, because if $E_{below}$ is less than zero, the actual E vector is pointing the other way.  That arrow there is just defining a direction, like $\hat{x}$, $\hat{y}$ and $\hat{z}$.
A: To support that Griffiths is correct, I would start with quoting from Bob Jacobsen's answer:

He’s picked a sign convention where the field upward is positive everywhere. That means that $E_{below}$ is defined such that a positive value means "the E vector points up" and a negative value means "the E vector points down".

Adding to this, I would also consider the convention that the direction of the area vector is positive along the upward direction and vice-versa.
Now, let's start from Gauss' law which says that $\boxed{Q_{enc}=\Phi\cdot\epsilon_0}$, where $Q_{enc}$ is the charge enclosed by the Gaussian pillbox and $\Phi$ is the electric flux.
Applying this to the Gaussian pillbox in the figure, we can write
\begin{align*}
Q_{enc} &=\epsilon_0\left[(E_{above}^\perp)\cdot(A)+(E_{below}^\perp)\cdot(-A)\right]\\ \implies\sigma A&=\epsilon_0A\left[E_{above}^\perp-E_{below}^\perp\right]\\ \implies\frac{\sigma}{\epsilon_0}&=E_{above}^\perp-E_{below}^\perp
\end{align*}
Note that the area vector along $E_{above}^\perp$ is positive and the area vector along $E_{below}^\perp$ is negative due to the chosen convention.
