I think the main thing to notice is that the expression
$$e_D(f) = f^{\frac{1}{D}}$$
is really really steep at the beginning. This means that the size of the edge that you will need to encompass a certain fraction of the volume will increase drastically, specially at the beginning. i.e. the edge you need will become ridiculously large as $D$ increases.
To make this even clearer, recall the plot that Murphy shows:
if you notice, for values of $D > 1$, the slope is really large and hence, the function grows really steeply at the beginning. This can be better appreciated if you take the derivative of $e_D(f)$:
$$ e'_D(f) = \frac{1}{D} f^{\frac{1}{D} - 1} = \frac{1}{D} f^{\frac{1 - D}{D}} $$
Since we are only considering increasing dimension (that are integer values), we only care for integer values of $D > 1$. This means that $1-D < 0$. Consider the expression for the edge as follows:
$$ e'_D(f) = \frac{1}{D} (f^{1 - D})^{\frac{1}{D}} $$
Notices that we are raising $f$ to a power less than 0 (i.e. negative). When we raise number to negative powers we are at some point doing a reciprocal (i.e. $x^{-1} = \frac{1}{x}$). Doing a reciprocal to a number that is already really small ( recall $f < 1$ since we are only considering fraction of the volume, since we are doing KNN, i.e. $k$ nearest data points out of the total $N$) means that number will "grows a lot". Therefore, we get the desired behavior, i.e. that as $D$ increases the power becomes even more negative and hence, the edge required grows a lot depending how large $D$ increases the exponent.
(notice that $f^{1 - D}$ grows exponentially compared to the division $\frac{1}{D}$ that quickly becomes insignificant).