Let $I\in\mathbb{R}^{d\times d}$ be the identity matrix. Let $x_{1},x_{2}\ldots,x_{N}$ and $y_{1},y_{2}\ldots,y_{N}$ be vectors in $\mathbb{R}^{d\times1}$.
Is there some expression for
$\mathrm{det}\left(I+x_{1}y_{1}^{\top}+\ldots+x_{N}y_{N}^{\top}\right)?$
Sylvester's determinant identity (also called the Weinstein-Aronszajn identity) allows you to rewrite this in a way that is particularly useful when $N < d$. Let $X$ denote the matrix whose columns are $x_1,\dots,x_N$ and $Y$ the matrix whose columns are $y_1,\dots,y_N$. Then the determinant you're after is $\det(I + XY^T)$, and we have $$ \det(I_d + XY^T) = \det(I_N + Y^TX). $$ In terms of the original vectors, this can be written as $$ \det(I_d + x_1y_1^\top + \cdots + x_Ny_N^\top) = \det \left[I_N + \pmatrix{y_1^\top x_1 & \cdots & y_1^\top x_N\\ \vdots & \ddots & \vdots\\ y_N^\top x_1 & \cdots & y_N^\top x_N} \right]. $$ Of course, this determinant can be written out as a polynomial over the dot-products $y_i^\top x_j$ if that is desired, but I'd say that this is impractical for $N > 3$.