Why is $(\sqrt{2}+\sqrt{3})^{2008}$ so close to an integer?

That is because

$$(\sqrt{3}+\sqrt{2})^{2m} + (\sqrt{3}-\sqrt{2})^{2m}$$

is an integer, namely

$$\begin{align} (\sqrt{3}+\sqrt{2})^{2m} + (\sqrt{3}-\sqrt{2})^{2m} = 2\sum_{k=0}^{m} \binom{2m}{2k} 3^{m-k}2^k, \end{align}$$

and $(\sqrt{3}-\sqrt{2})^{2m} <\frac{1}{3^{2m}}$.

Your intuition is correct: $$ (\sqrt{2} + \sqrt{3})^{2008} = (5 + 2 \sqrt{6})^{1004} $$ and $5 + 2\sqrt{6}$ is a Pisot integer. Further, Newton's identities imply that $$ (5 + 2\sqrt{6})^n + (5 - 2\sqrt{6})^n \in \Bbb{Z} $$ for every positive integer $n$, thus the distance between $(5 + 2\sqrt{6})^{1004}$ and the closest integer is at most $$ (5 - 2\sqrt{6})^{1004} \approx 2.6743 \cdot 10^{-1000} $$