How to calculate the determinant of all-ones matrix minus the identity?
$$D_n(a,b)= \begin{vmatrix} a & b & b & b \\ b & a & b & b \\ b & b & a & b \\ b & b & b & a \end{vmatrix}$$ ($n\times n$-matrix).
$$D_n(a,b)= \begin{vmatrix} a & b & b & b \\ b & a & b & b \\ b & b & a & b \\ b & b & b & a \end{vmatrix}$$
$$=[a+(n-1)b] \begin{vmatrix} 1 & 1 & 1 & 1 \\ b & a & b & b \\ b & b & a & b \\ b & b & b & a \end{vmatrix}$$ $$=[a+(n-1)b] \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & a-b & 0 & 0 \\ 0 & 0 & a-b & 0 \\ 0 & 0 & 0 & a-b \end{vmatrix}$$ $$=[a+(n-1)b](a-b)^{n-1} $$
(In the first step we added the remaining rows to the first row and then "pulled out" constant out of the determinant. Then we subtracted $b$-multiple of the first row from each of the remaining rows.)
You're asking about $D_n(0,1)=(-1)^{n-1}(n-1)$.
If you replace one column by 1's, you can use this result to get the following. (I've computed it for $n=4$, but I guess you can generalize this for arbitrary $n$.)
$$ \begin{vmatrix} 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{vmatrix} = \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{vmatrix} + \begin{vmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{vmatrix}= \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{vmatrix} + \begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} $$
Note that both these determinants are of the type you already handled in the first part.
Here's an approach using Sylvester's determinant theorem, which says that for any rectangular matrices of mutually transposed shapes $A\in\mathrm M_{n,m}(K)$ and $B\in \mathrm M_{m,n}(K)$ one has $$\det(I_n+AB)=\det(I_m+BA).$$
If $N$ is your matrix then $-N=I_n-AB$ where $A\in\mathrm M_{n,1}(K)$ is a one column all-one matrix and $B$ is its transpose. Then $$ \det(N)=(-1)^n\det(-N)=(-1)^n\det(I_1-BA)=(-1)^n(1-n). $$
I will compute the determinant of the matrix $$ A = \left( \begin {matrix} b & a & \ldots & a \\ a & b & \ddots & \vdots \\ \vdots & \ddots & \ddots & a \\ a & \cdots & a & b \end {matrix} \right), $$ where $a, b \in \mathbb{K}$. To obtain your case, put $a=1$ and $b=0$.
First proof. This works if $\mathrm{char}(\mathbb{K}) = 0$ or $n$ is prime to $\mathrm{char}(\mathbb{K}) > 0$. If $a =0$, then $\det A = b^n$. Suppose $a \neq 0$ and consider the vector $v = (1, \dots, 1) \in \mathbb{K}^n$; it is clear that $v$ is an eigenvector of $A$ with eigenvalue $\alpha = (n-1) a + b$. Now consider $\beta = b-a$. $\beta$ is an eigenvalue of $A$ because $$ B = A - \beta I_n = \left( \begin {matrix} a & a & \ldots & a \\ a & a & \ldots & a \\ \vdots & \vdots & \ddots & \vdots \\ a & a & ... & a \end {matrix} \right) $$ has rank $1$.
Let $E_\alpha, E_\beta \subseteq \mathbb{K}^n$ the eigenspaces of $A$ of the eigenvalues $\alpha, \beta$. We have $\alpha \neq \beta$, $E_\alpha \cap E_\beta = \{ 0 \}$ and $\dim E_\beta = n-1$, thus $\mathbb{K}^n = E_\alpha \oplus E_\beta$ and $E_\alpha = \langle v \rangle$. This proves that $A$ is similar to the matrix $\mathrm{diag}(\alpha, \beta, \dots, \beta)$, therefore $\det A = \alpha \beta^{n-1} = [(n-1)a +b] (b-a)^{n-1}$. (Notice that this formula holds also when $a=0$.)
Second proof. The characteristic polynomial of the matrix $B = A - (b-a) I$ is $$ \chi_B(t) = (-t)^n + c_{n-1} (-t)^{n-1} + \cdots + c_1(-t) + c_0, $$ where $c_i$ is the sum of the principal $(n-i)$-minors of $B$. It is clear that all principal minors of $B$ are zero, except on $1$-minors. Thus $$ \chi_B(t) = (-t)^n + na (-t)^{n-1}. $$ From $A = B + (b-a) I$, we have $\chi_A(t) = \chi_B(t-b+a)$. Thus $\det A = \chi_A(0) = \chi_B(a-b) = (b-a)^n + na (b-a)^{n-1}$.
Now consider the matrix $$ C = \left( \begin {matrix} a & a & \ldots & a \\ a & b & \ddots & \vdots \\ \vdots & \ddots & \ddots & a \\ a & \cdots & a & b \end {matrix} \right). $$ For every $i=2,\dots,n$, replace the $i$th row $C_i$ of $C$ with $C_i - C_1$, where $C_1$ is the first row of $C$. Obtain $$ \det C = \det \left( \begin{matrix} a & a & a & \cdots & a \\ 0 & b-a & 0 & \cdots & 0 \\ 0 & 0 & b-a & \ddots & 0 \\ \vdots & \vdots & \ddots & \ddots & 0 \\ 0 & 0 & \cdots & 0 & b-a \end{matrix} \right) = a (b-a)^{n-1}. $$