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.
You have $\mathrm{tr}[ X \rho] =1$ for all $\rho$ where $X$ and $\rho$ are both positive semidefinite matrices.
Now as $X$ is PSD we know by the spectral theorem that
$$
X= U D U^\dagger = \sum_i \lambda_i |u_i\rangle \langle u_i|
$$
where $\lambda_i$ are the nonnegative eigenvalues of $X$ and $|u_i\rangle$ are the corresponding orthonormal eigenvectors. Note that $X = I$ if we can show that each $\lambda_i =1$. But this follows from the condition above by taking $\rho = |u_i\rangle \langle u_i|$ we have
$$
\mathrm{tr}[ X |u_i\rangle \langle u_i|]=1 \implies \lambda_i = 1
$$
Repeating for each eigenvector we see that $X = U I U^\dagger = U U^\dagger = I$.
Best Answer
There are a number of things wrong with your computation.
The expression "$\operatorname{Tr}_{\mathcal{F}}(\rho)$" doesn't make any sense, as $\rho$ is an operator on $\mathcal{N}$ and not on $\mathcal{N}\otimes\mathcal{F}$. The expressions "$\rho\phi_{i}$" likewise are meaningless, as $\phi_i\in\mathcal{F}$.
The partial trace is not invariant under cyclic permutations, so your first step is incorrect.
The operator $V^\dagger V\rho$ isn't square (it is an operator from $\mathcal{N}\rightarrow\mathcal{N\otimes\mathcal{F}}$) so it doesn't make sense to take its partial trace.
While the trace of an operator $X\in \operatorname{L} (\mathcal{F})$ is computed as $\operatorname{Tr}(X)=\sum_{i}\langle \phi_i, X\phi_i\rangle$, the partial trace is not computed this way. Indeed, if $X\in \operatorname{L} (\mathcal{N\otimes F})$, the expression "$\langle \phi_i, X\phi_i\rangle$" doesn't even make sense.
In general, the expression "$\operatorname{Tr}_\mathcal{F}(V\rho V^\dagger)$" may not be simplified further.