Replace the first column by the sum of all the columns: the wanted determinant is equal to
$$\Delta_n:=\det\begin{pmatrix}
1 & -1 & \cdots & -1 \\
1 & n-1 & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots \\
1 & \cdots & \cdots & n-1
\end{pmatrix}.$$
Now, for each $j\in\{2,\dots,n\}$, take the column $C_j$ and replace it by $C_j+C_1$ to obtain
$$\Delta_n=\det\begin{pmatrix}
1 & 0 & \cdots & 0 \\
1 & n & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots \\
1 & 0 & \cdots & n
\end{pmatrix}.$$
The determinant of a triangular matrix is easier to compute.
I would like to present a very simple solution by interpretation of these matrices as operators on $\mathbb{R^n}$ (which will surprise nobody...). Triangular matrix $A$ acts as a discrete integration operator:
For any $x_1,x_2,x_3,\cdots x_n$:
$$\tag{1}A (x_1,x_2,x_3,\cdots x_n)^T=(s_1,s_2,s_3,\cdots s_n)^T \ \ \text{with} \ \ \begin{cases}s_1&=&x_1&&&&\\s_2&=&x_1+x_2&&\\s_3&=&x_1+x_2+x_3\\...\end{cases}$$
(1) is equivalent to:
$$\tag{2}A^{-1} (s_1,s_2,s_3,\cdots x_n)^T=(x_1,x_2,x_3,\cdots x_n)^T \ \ \text{with} \ \ \begin{cases}x_1&=& \ \ s_1&&&&\\x_2&=&-s_1&+&s_2&&\\x_3&=&&&-s_2&+&s_3\\...\end{cases}$$
and it suffices now to "collect the coefficients" in the right order in order to constitute the inverse matrix.
(Thus the inverse operation is - in a natural way - a discrete derivation operator).
Best Answer
$$ {\bf M}_n (a) := \begin{bmatrix} 1 & a & a & \dots & a & a\\ a & 1 & a & \dots & a & a\\ a & a & 1 & \dots & a & a\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a & a & a & \dots & 1 & a\\ a & a & a & \dots & a & 1\end{bmatrix} = (1-a) {\bf I}_n + a {\bf 1}_n {\bf 1}_n^{\top} $$
Using Sherman-Morrison,
$$ {\bf M}_n^{-1} (a) = \cdots = \color{blue}{\frac{1}{1 - a} \left( {\bf I}_n - \frac{a}{1 + (n-1) a} {\bf 1}_n {\bf 1}_n^{\top} \right)} $$
which is the matrix that Greg obtained via other means.