What is true is that the expansion of the characteristic polynomials is given by traces of powers of the matrix $A$; explicitly, the characteristic polynomial $\chi_A(T)=\det(T\cdot{\rm Id}-A)$ is given by
$$T^n-{\rm tr}(A)T^{n-1}+\frac{{\rm tr}(A)^2-{\rm tr}(A^2)}{2}T^{n-2}-\frac{{\rm tr}(A)^3-3{\rm tr}(A){\rm tr}(A^2)+2{\rm tr}(A^3)}{6}T^{n-3}+\cdots $$
with the final term being $(-1)^n\det A$. This does not depend on $A$ being symmetric or anything else and so $\det(1+A)=(-1)^n\chi_A(-1)\approx (-1)^n[(-1)^n-{\rm tr}(A)(-1)^{n-1}]=1+{\rm tr}(A)$ is in some sense a valid first-order approximation (we set $T=-1$ and truncated our expansion at just two terms).
This can also be seen more simply as in Alex's answer: if $\{\lambda_i\}$ are the eigenvalues of $A$ with multiplicity, then $I+A$ has eigenvalues $\{\lambda_i+1\}$ with multiplicity, and $$\det(1+A)=\prod_{i=1}^n(1+\lambda_i)=1+\left(\sum_{i=1}^n\lambda_i\right)+{\rm higher~order~terms}\approx 1+{\rm tr}(A).$$
I have been assume $A$ is an $n\times n$ matrix throughout, but the underlying field is arbitrary.
The $\chi_A(T)$ expansion is possible thanks to the fundamental theorem of symmetric polynomials, and in particular may be recursively computed by hand using the NG formulas (see this section).