I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?
This is the matrix :
$$\begin{bmatrix}n&n-1&n-2&\cdots&2&1\\1&n&n-1&\cdots&3&2\\1&1&n&\cdots&4&3\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\1&1&1&\cdots&n&n-1\\1&1&1&\cdots&1&\lambda\end{bmatrix}$$
On
Let $M_n$ be your matrix.
Let $\eta_n$ be the $n\times n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you
Subtract row $k+1$ from row $k$ for $k = 1,2,\ldots,n-1$.
This is equivalent to multiply $M_n$ by $I_n - \eta_n$ from the left
Subtract column $k-1$ from column $k$ for $k = n,n-1,\ldots,2$ (notice the order of $k$).
This is equivalent to multiply $(I_n-\eta_n)M_n$ by $I_n - \eta_n$ from the right.
After you do this, your matrix simplifies to $$(I_n - \eta_n) M_n (I_n - \eta_n) = \begin{bmatrix} n-1&-n&0&\cdots&0&0&0\\ 0&n-1&-n&\cdots&0&0&0\\ 0&0&n-1&\ddots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&n-1&-n&0\\ 0&0&0&\cdots&0&n-1&-\lambda\\ 1&0&0&\cdots&0&0&\lambda-1 \end{bmatrix}$$
From this, you can deduce
$$\det[M_n] = \det[(I_n - \eta_n)M_n(I_n - \eta_n)] = (n-1)^{n-1}(\lambda-1) + n^{n-2}\lambda$$
This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.
Then $\det(A) = \det(B)(1+ e^TB^{-1}e)$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.
This is defined if $\lambda \ne 1$. For $\lambda = 1$, take the limit.