\begin{vmatrix}
0 & 2 & 3 & 15 \\
2 & 0 & 5 & 9 \\
3 & 5 & 0 & 6 \\
15 & 9 & 6 & 0
\end{vmatrix}
This matrix has determinant zero.
Added: The determinant in the $4\times 4$ case is quite nice:
$$2[(af)^2+(be)^2+(cd)^2]-(af+be+cd)^2. $$ To get integer entries, consider
the Diophantine equation $$2[x^2+y^2+z^2]=(x+y+z)^2 $$
which has solutions $x=cr^2$, $y=cs^2$, $z=ct^2$ where $r+s=t$. We need to be
able to split $x,y,z$ into six distinct factors, two each. I eventually hit upon $c=3$, $r=2$, $s=3$, and $t=5$. This gives $x=12$, $y=27$, $z=75$ which I split as $a=2$, $f=6$, $b=3$, $e=9$, $c=15$, and $d=5$.
This determinant measures the volume of the parallelepiped built on the column vectors. If you keep the vector lengths constant, which keeps the sum of squares constant, the volume is maximized when the vectors are orthogonal.
Now let the lengths be $a,b,c$, you want to maximize
$$V=abc$$ under the constraint $$a^2+b^2+c^2=1.$$
By Lagrangian multipliers,
$$abc+\lambda(a^2+b^2+c^2-1)$$
$$\begin{cases}bc+2\lambda a&=0,
\\ac+2\lambda b&=0,
\\ab+2\lambda c&=0,
\\a^2+b^2+c^2-1&=0\end{cases}$$
Eliminating $\lambda$ from the three first equations, you get $a^2=b^2=c^2$, and using the last, $a^2=b^2=c^2=\dfrac13$. Hence
$$\Delta=\frac1{\sqrt{27}}.$$
Direct vector proof:
Let $\vec a,\vec b,\vec c$ be the column vectors of the matrix. The determinant is the mixed product
$$\Delta=\vec a\cdot(\vec b\times\vec c).$$
Using a Lagrangian multiplier, we maximize
$$\vec a\cdot(\vec b\times\vec c)+\lambda(\vec a^2+\vec b^2+\vec c^2-1).$$
Taking the gradient yields
$$\begin{cases}\vec b\times\vec c+2\lambda\vec a=0,
\\\vec c\times\vec a+2\lambda\vec b=0,
\\\vec a\times\vec b+2\lambda\vec c=0,
\\\vec a^2+\vec b^2+\vec c^2-1=0.
\end{cases}$$
Then applying dot products on the first equations, we get
$$\lambda\vec a\vec b=\lambda\vec b\vec c=\lambda\vec c\vec a=0,$$ which shows that the vectors must be orthogonal, and we are back to the above scalar problem.
Best Answer
Let the matrix be \begin{align} A & = \begin{bmatrix} d_1 & 1 & 1 & \cdots & 1\\ 1 & d_2 & 1 & \cdots & 1\\ 1 & 1 & d_3 & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & d_n \end{bmatrix}\\ & = \begin{bmatrix} d_1-1 & 0 & 0 & \cdots & 0\\ 0 & d_2-1 & 1 & \cdots & 0\\ 0 & 0 & d_3-1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & d_n-1 \end{bmatrix}+ \begin{bmatrix} 1\\ 1\\ 1\\ \vdots\\ 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \end{bmatrix}\\ & = D+uu^T \end{align} where $D=\begin{bmatrix} d_1-1 & 0 & 0 & \cdots & 0\\ 0 & d_2-1 & 1 & \cdots & 0\\ 0 & 0 & d_3-1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & d_n-1 \end{bmatrix}$ and $u^T=\begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \end{bmatrix}$.
From Sylvester determinant theorem, we have $$\det(D+uu^T) = \det(D)\det(I+D^{-1}uu^T) = \det(D) \det(I_{1 \times 1} + u^TD^{-1}u)$$
This gives us that the determinant is $$\prod_{k=1}^n (d_k-1)\left(1+\sum_{i=1}^n\dfrac1{d_i-1}\right) = \prod_{k=1}^n (d_k-1) + \sum_{i=1}^n \prod_{\overset{k=1}{k \neq i}}^n (d_k-1)$$