The Fredholm determinant is not sequentially continuous in the strong topology. Take the sequence of 1-dimensional projectors $X_n=\langle e_n,\cdot\rangle e_n$ with the $e_n$ forming an ON basis. You then have $X_n \to 0$ strongly, but $0=\det(I-X_n)$ does not converge to $1=\det(I)$. Or what did you have in mind when saying that you would know the continuity for the strong topology (once restricted to trace class)?
If you, however, add the convergence $\|X_n\|_1\to\|X\|_1$, where $\|\cdot\|_1$ denotes trace class norm, then it is known that weak sequential convergence $X_n \to X$ implies convergence in trace class norm and, therefore, of the Fredholm determinant (see Thm. 2.21 and Addendum H of the book "Trace Ideals and Their Applications" by Barry Simon, 2nd ed., AMS 2005).
The Fredholm determinant is not sequentially continuous in the strong topology. Take the sequence of 1-dimensional projectors $X_n=\langle e_n,\cdot\rangle e_n$ with the $e_n$ forming an ON basis. You then have $X_n \to 0$ strongly, but $0=\det(I-X_n)$ does not converge to $1=\det(I)$. Or what did you have in mind when saying that you would know the continuity for the strong topology (once restricted to trace class)?