Once the confusion between $\mathbb E(|X_n|)$ (the mean distance from the origin) and $|\mathbb E(X_n)|$ (the absolute value of the mean position) is cleared, it seems a second kind of confusion might explain your trouble. There are two very different cases. Either $p\ne q$, then $X_n\sim n(p-q)$ almost surely and both $|\mathbb E(X_n)|$ and $\mathbb E(|X_n|)$ are asymptotically $n|p-q|$ (then one says the walk is ballistic). Or $p=q=\frac12$, then $X_n$ behaves like $\sqrt{n}$ times a standard normal random variable and $|\mathbb E(X_n)|=0$ for every $n$ while $\mathbb E(|X_n|)\sim\sqrt{n}$ (then one says the walk is diffusive).

Once the confusion between $\mathbb E(|X_n|)$ (the mean distance from the origin) and $|\mathbb E(X_n)|$ (the absolute value of the mean position) is cleared, it seems a second kind of confusion might explain your trouble. There are two very different cases. 

Either $p\ne q$, then $X_n\sim n(p-q)$ almost surely and both $|\mathbb E(X_n)|$ and $\mathbb E(|X_n|)$ are asymptotically $n|p-q|$ (then one says the walk is ballistic). Or $p=q=\frac12$, then $X_n$ behaves like $\sqrt{n}$ times a standard normal random variable and $|\mathbb E(X_n)|=0$ for every $n$ while $\mathbb E(|X_n|)\sim\sqrt{n}$ (then one says the walk is diffusive).

A more usual formulation of the diffusive property uses the root mean square displacement $\sqrt{\mathbb E(X_n^2)}$ and states that $\mathbb E(X_n^2)\sim n$. The nearly universal use of the root mean square displacement rather than the mean displacement is mainly due to computational feasibility, the computation of the mean square displacement being often easy while the computation of the mean displacement is often impossible.