For orders $n > 3$, these matrices have a zero determinant because they always have a non zero vector in their kernel. Moreover, this vector is the same for all matrices with the same order:
$$\tag{1} \ \ \ \begin{pmatrix}\ \ 1\\-3\\ \ \ 3\\-1\end{pmatrix}, \ \ \ \ \begin{pmatrix}\ \ 1\\-4\\ \ \ 6\\-4 \\ \ \ 1\end{pmatrix}, \ \ \text{etc.}$$
depending if $n=4, 5, \cdots$ etc.
(more generally, a vector whose entries are the coefficients of $(1-x)^{n-1}$.)
This is not difficult to prove; for example in the $4 \times 4$ case, it is based on the identity:
$$\tag{2}\forall n \in \mathbb{Z}, \ \ n^2-3(n+1)^2+3(n+2)^2-(n+3)^2=0$$
The general case being almost as simple.
As you are clearly interested by understanding what is "behind the curtain" there is a deeper reason than somewhat "accidental" algebraic identities like in $(2)$.
I had the idea of using these vectors because I was rather familiar with finite differences calculus, which is parallel, up to a certain point, with "continuous" (differential) calculus. Have a look for example at (http://mathworld.wolfram.com/FiniteDifference.html).
In short: as we work with second degree expressions, multiplying by "masks" with $(1, \ \ -3, \ \ 3, \ \ -1)$ or $(1, \ \ -4, \ \ 6, \ \ -4, \ \ 1)$ coefficients amounts to differentiate three times, four times, etc... a second degree expression, "naturally" giving a zero result.
Remark 1: The null determinant property can be extended to a larger class of matrices, i.e., matrices having consecutive squares entries in a same row, but with a possible disruption with the next row, for example:
$$\begin{pmatrix}\ \ 1^2&2^2&3^2&4^2\\ 7^2&8^2&9^2&10^2\\ 3^2&4^2&5^2&6^2\\11^1&12^2&13^2&14^2\end{pmatrix}$$
Remark 2: I think that, beyond the fact that the determinant is $0$, a more accurate characterization is through the rank of the matrix. let us understand it on the example of $SMCS(5,7)$.
Let us denote by $M,V_1,V_2$ and $V_3$ the following matrix and vectors:
$$M=\begin{pmatrix}\ 49 & 64 & 81 & 100 & 121\\
144 & 169 & 196 & 225 & 256\\
289 & 324 & 361 & 400 & 441\\
484 & 529 & 576 & 625 & 676\\
729 & 784 & 841 & 900 & 961\end{pmatrix} \ \ \text{and} \ \ \ V_1=\begin{pmatrix}\ \ 1\\-3\\ \ \ 3\\-1 \\ \ \ 0\end{pmatrix}, V_2=\begin{pmatrix} \ \ 0 \\ \ \ 1\\-3\\ \ \ 3\\-1\end{pmatrix}, V_3=\begin{pmatrix}\ \ 1\\-4\\ \ \ 6\\-4 \\ \ \ 1\end{pmatrix}$$
(note that $V_1$ and $V_2$ are made with the first vector of (1), with a n added $0$ at the and or at the beginning.)
$\det(M)=0$ but with a more important rank loss than expected: rank$(M)=3=5-2$. It means that the kernel has dimension 2. It is interesting to see that a basis of the kernel is constituted by $V_1$ and $V_2$, with $V_3=V_1-V_2$ which is easily explained (the mask of the fourth discrete differentiation operator is the difference of the masks of the third discrete diff. operator).
If $h_k$ are the projections of $p_k$ onto $x$ axis, then we have:
$$
A(p_1,p_2,p_3)=A(h_1,h_3,p_3,p_1)-A(h_1,h_2,p_2,p_1)-A(h_2,h_3,p_3,p_2),
$$
where I assumed $x_1<x_2<x_3$ to avoid using absolute values.
The areas of the three trapezoids on the r.h.s. can be computed and the result is:
$$
A(p_1,p_2,p_3)=
-{1\over2}\big[(x_1^2+x_3^2)(x_1-x_3)+(x_1^2+x_2^2)(x_2-x_1)
+(x_2^2+x_3^2)(x_3-x_2)\big],
$$
which is the same as the expression given in the question.
Best Answer
Assume none of the eigenvalues of $A$ lie on the negative real axis $(-\infty,0]$, it is always possible to define $A^s$ in such a way that $\det(A^s) = \det(A)^s$. To see that, use a similarity transform to bring $A$ into its Jordan normal form and works with it.
If $A$ has zero as an eigenvalue, then one may get into trouble in defining $A^s$.
The simplest example I can think of is $A = \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}$.
If all eigenvalues of $A$ are non-zero but some of them are negative, we can define $A^s$ but we may get into problem to enforce $\det(A^s) = \det(A)^s$. As an example, consider
$$A = \begin{bmatrix}-1 & 0\\ 1 & -1\end{bmatrix}$$
We have $\det(A) = 1$ but any sensible/natural definition of $A^s$ will force $\det(A^s) = e^{2\pi n s i}$ for some odd integer $n$.