$\DeclareMathOperator{\tr}{tr}$ The exact expansion is $$det(1+tA) = \sum t^i \sigma_i(\lambda_1, \ldots, \lambda_n),$$ where $\lambda_i$ are the eigenvalues of $A$ and $\sigma_i$ is the $i$-th elementary symmetric polynomial, e.g. $\sigma_0=1, \sigma_1=\sum \lambda_i, \sigma_2=\sum _{i\ne j}\lambda_i \lambda_j, \ldots$ These are the same coefficients as in the characteristic polynomial of $A$, but in reverse order (perhaps up to sign)

As pointed out in the comments, there are ways to get these coefficients from traces of powers of your matrix, if these are any easier to come by than the eigenvalues themselves; this is because $\tr(A^k)=\sum \lambda_i^k$ forms another basis (at least rationally) of the ring of symmetric polynomials in the $\lambda_i$, so one needs only to determine the base change coefficients between this basis and the one consisting of the $\sigma_i$.

(Newton's identitites are one explicit way to do this, there may be better)

$\DeclareMathOperator{\tr}{tr}$ The exact expansion is $$\det(1+tA) = \sum t^i \sigma_i(\lambda_1, \ldots, \lambda_n),$$ where $\lambda_i$ are the eigenvalues of $A$ and $\sigma_i$ is the $i$-th elementary symmetric polynomial, e.g. $\sigma_0=1, \sigma_1=\sum \lambda_i, \sigma_2=\sum _{i\ne j}\lambda_i \lambda_j, \ldots$ These are the same coefficients as in the characteristic polynomial of $A$, but in reverse order (perhaps up to sign)

As pointed out in the comments, there are ways to get these coefficients from traces of powers of your matrix, if these are any easier to come by than the eigenvalues themselves; this is because $\tr(A^k)=\sum \lambda_i^k$ forms another basis (at least rationally) of the ring of symmetric polynomials in the $\lambda_i$, so one needs only to determine the base change coefficients between this basis and the one consisting of the $\sigma_i$.

(Newton's identitites are one explicit way to do this, there may be better)