I was thinking of an example of a matrix $A$ which is at least $4$ dimensional, that has no real eigenvalue but yet det$(\Phi) = 0$, where the matrix $\Phi$ is given by $\begin{bmatrix} e_{1}^{T} & \\ e_{1}^{T}A \\ e_{1}^{T}A^2 \\ .\\ .\\ .\\ e_{1}^{T}A^{n-1}\end{bmatrix}$
Where $e_{1}$ is the standard basis vector of first entry 1 and others 0.
The eigenvalues of $A$ are complex so they must occur in pairs and hence I think we, therefore, need for $A$ to be even-dimensional and must be greater than or equal to $4$, also the determinant is the product of the eigenvalues will be a positive number then as we will be multiplying the complex eigenvalue and its conjugate, so the determinant will be positive but I am thinking now how the determinant can be zero? as there is no eigenvalue 0.
The determinant asked about is not that of $A$, but that of repeated (right) images of the first row $e_1^T$ of the identity matrix. If you take $A$ to be a block diagonal matrix, then those images remain "inside the first block", so they will never produce a full rank matrix, and the determinant will be$~0$. So take a block diagonal matrix of at least two $2\times 2$ rotation matrices, which have no real eigenvalues, like $$ A=\pmatrix{0&1&0&0\\-1&0&0&0\\0&0&0&-1\\0&0&1&1} $$ which gives $$ \Phi=\pmatrix{1&0&0&0\\0&1&0&0\\-1&0&0&0\\0&-1&0&0}, $$ obviously singular.