Edit: years after I posted this answer I wrote a proof for some notes I was preparing, so now I'm posting the proof here. I left the old answer below the new proof.
Because any reordering of an orthonormal basis is an orthonormal basis, the hypothesis is equivalent to requiring absolute convergence (a series of complex numbers converges under every possible reordering if and only if it converges absolutely). Also, since $\def\Re{\operatorname{Re}}\def\Im{\operatorname{Im}}\def\e{\varepsilon}\def\RR{\mathbb R}$
$$
\Re\sum_n\langle T\xi_n,\xi_n\rangle
=\sum_n\Re\langle T\xi_n,\xi_n\rangle
=\sum_n\langle \Re T\xi_n,\xi_n\rangle
$$
and similarly for the imaginary parts, the operators $\Re T$ and $\Im T$ also satisfy the hypothesis. As linear combinations of trace-class are trace-class, we may assume without loss of generality that $T$ is selfadjoint.
If $T$ is not compact, via the Spectral Theorem there exists $\lambda\in\sigma(T)\setminus\{0\}$ and $\delta>0$, with either $\lambda-\delta>0$ or $\lambda+\delta<0$, such that the spectral projection $P=\mu_T(\lambda-\delta,\lambda+\delta)$ is infinite. With $\{\xi_n\}$ an orthonormal basis for the range of $P$ we have, with $\e=\min\{|\lambda-\delta|,|\lambda+\delta|\}$,
$$
\sum_n|\langle T\xi_n,\xi_n\rangle|\geq\sum_n\e\,\langle\xi_n,\xi_n\rangle=\sum_n\e=\infty,
$$
contradicting the hypothesis. So $T$ is compact. Now the Spectral Theorem for Compact Operators allows us to write
$$
T=\sum_n\lambda_nP_n,
$$
where $\{\lambda_n\}\subset\RR\setminus\{0\}$ is a sequence that converges to $0$, and $\{P_n\}$ are pairwise orthogonal rank-one projections. If $T$ is not trace-class, this means that
$$
\sum_n|\lambda_n|=\infty.
$$
Let $\{\xi_n\}$ be unit vectors with $P_n\xi_n=\xi_n$. Then $\{\xi_n\}$ is orthonormal and
$$
\sum_n|\langle T\xi_n,\xi_n\rangle|=\sum_n|\lambda_n|=\infty.
$$
This contradicts the hypothesis, so the only possibility is that $T$ is trace-class.
(original answer below)
It's not a dumb question; actually, it's been asked on this site more than once, and never answered. And I cannot give you a full answer because I don't know it, but here are some thoughts.
The condition works when $A$ is positive, and in that case convergence for only one orthonormal basis is sufficient, and it agrees with the "good" definition. Maybe that's what your source is doing?
But when $A$ is not positive, you cannot initially say that the value of $\sum_n\langle Ae_n,e_n\rangle$ does not depend on the orthonormal basis, because to prove that you need to exchange series, and for that you need absolute convergence.
Absolute convergence does apply, though, because you are allowed to reorder any fixed basis. And a series converges under all permutations if and only if it is absolutely convergent.
The problem is that you still don't know that $\sum_n\langle Ae_n,e_n\rangle$ convergent (note that there is no need for the value to be positive or even real), implies that the sum is the same for all bases. The usual argument goes by showing that $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$ and then using it to show that $\operatorname{Tr}(V^*AV)=\operatorname{Tr}(A)$ for any unitary, but this requires $AB$ and $BA$ to be trace-class. With the usual definition that $\sum_n\langle (A^*A)^{1/2}e_n,e_n\rangle<\infty$, one has inequalities available that allow you to show that the space of trace-class operators is an ideal and so if $A$ is trace-class so are $AB$ and $BA$. But with your definition, I don't see how you could show this.
In summary, I cannot prove that your definition is wrong (that would require finding an operator such that those sums are finite for all orthonormal bases while the operator is not trace-class), but at the very least it is not useful unless some smart calculation allows you to use it to show that the trace-class operators, as you define them, form an ideal.
For more information on this, here you can see how a stronger requirement than yours does imply trace-class.
