I'd recommend first that you and your friend spend more time with Tits :), "Reductive groups over local fields", from the Corvallis volume (free online, last time I checked). Undoubtably there are other references, like papers of Prasad-Raghunathan mentioned by Greg Kuperberg, and any paper that treats Bruhat-Tits theory for unitary groups. I'll try to provide a background/basic treatment here.

You're certainly used to Bruhat-Tits theory for $SL_2$, or at least for $PGL_2$, over $Q_p$, where one encounters the $(p+1)$-regular tree. As you know, $PGL_2(Q_p)$ acts on this tree, and the stabilizer of a point is a maximal compact subgroup that is conjugate to $PGL_2(Z_p)$, and the stabilizer of an edge is an Iwahori subgroup. This I assume is a familiar picture.

To understand the "special point" subtleties, you should first think about a group like $G=SU_3$ -- the $Q_p$-points of a quasisplit form of $SL_3$, associated to an *unramified* quadratic field extension $K/Q_p$ with $p$ odd. Let $S$ be a maximal $Q_p$-split torus in $G$. Then $S$ has rank one, though the group $G$ has absolute rank two. It follows that the Bruhat-Tits building for $SU_3$ is again a tree, though not as simple as the $SL_2$ case. In fact, in this unramified situation, the building can be seen as the fixed points of the building of $G$ over $K$ (which is the building of $SL_3$), under the Galois involution.

Now, one must think about the *relative* roots of $G$ with respect to $S$ -- i.e., decompose the Lie algebra of $G$ with respect to the adjoint action of $S$. There are four eigenspaces with nontrivial eigenvalue -- these are the root spaces for the relative roots which I'll call $\pm \alpha$ and $\pm 2 \alpha$. The root spaces ${\mathfrak g}_{\pm \alpha}$ are two-dimensional.

Now, let $A$ be the apartment of the building associated to $S$ -- $A$ is a principal homogeneous space for the one-dimensional real vector space $X_\bullet(S) \otimes_Z R$. After choosing a good base point (a hyperspecial base point, using the fact that $K/Q_p$ is unramified), $A$ may be identified with $X_\bullet(S) \otimes_Z R$ and the affine roots are functions of the form $\pm \alpha + k$ and $\pm 2 \alpha + k$, where $k$ can be any integer.

Let $h$ be a generator of the rank 1 $Z$-module $X_\bullet(S)$, so that $\alpha^\vee = 2 h$, and $A = R \cdot h$. The affine roots are given by:
$$[\pm \alpha + k](r h) = r + k, [\pm 2 \alpha + k](r h) = 2r + k.$$
The vanishing hyperplanes of these affine roots are the points:
$$r h : r \in \frac{1}{2} Z.$$
These are the vertices of the building, contained in the apartment $A$.

Now consider a vertex $nh$, where $n$ is an integer. The affine roots $\pm (\alpha - n)$ and $\pm (2 \alpha - 2n)$ vanish at the vertex $n$. The *gradients* of these affine roots are the roots $\pm \alpha$ and $\pm 2 \alpha$. These are *all* of the roots in the original (relative) root system. That's why these vertices are hyperspecial vertices.

On the other hand, consider a vertex $(n + \frac{1}{2}) h$, where $n$ is an integer. The affine roots vanishing at this vertex are $\pm (2 \alpha - 2n - 1)$. The gradients of these affine roots are the roots $\pm 2 \alpha$. These are *not* all of the original roots, but all original roots are *proportional* to these roots. You can see how this phenomenon requires the setting of a non-reduced root system to happen. These "half-integral" vertices are special points, since the original root system does not occur in the system of gradients, but it does up to proportionality. At these special (but not hyperspecial) points, the Bruhat-Tits group scheme underlying the parahoric has special fibre with reductive quotient isomorphic to $PGL_2$ (I think... or is it $SL_2$) over the residue field. At the hyperspecial points, the group would be a quasisplit $SU_3$ over the residue field.

If it's not clear from above, a special point in the building occurs where the set of gradients of affine roots vanishing at that point is equal, modulo proportionality, to the set of relative roots. That's the general definition.

Hope this helps - see Tits for more.