Since $A$ is skew-symmetric, $a_{i,i}=-a_{i,i}$ so $a_{i,i}=0$.
Hence $Tr\;A=0$.
Now you have seen that $Tr\;B=5$.
So $Tr\;C=Tr\;A+Tr\;B=0+5=5$.
Let us start with another basis-independent yet more tractable (as it does not require the characteristic polynomial to split) definition of the trace. We will check in the end that it coincides with your definition, and with the sum of the diagonal coefficients with respect to any basis.
Let $V$ be an $n$-dimensional vector space over a field $F$. And let $L(V)$ be the algebra of $F$-linear maps from $V$ to $V$.
Note that we have a canonical isomorphism
$$
L(V)\simeq V\otimes V^*
$$
via $v\otimes w^* \simeq w^*(\cdot)v$. In other words, $L(V)$ is a natural incarnation of the tensor product of $V$ with its dual $V^*$, with rank-one operators as elementary tensors.
Observe that the bilinear map $(v,w^*)\longmapsto w^*(v)$ factors uniquely through the tensor product.
That's the trace, which is therefore characterized by
$$
\mathrm{tr}:V\otimes V^*\longrightarrow F\qquad \mathrm{tr}(v\otimes w^*)=w^*(v).
$$
Now choose any basis $\{e_i\}$ for $V$ and denote its dual basis by $\{e_i^*\}$. We have $\mathrm{tr}(e_i\otimes e_j^*)=\delta_{ij}$. Therefore, for every $x=\sum x_{ij}e_i\otimes e_j^*\in L(V)$, we have
$$
\mathrm{tr} (x)=\sum_{i=1}^n x_{ii}.
$$
Conclusion When given a matrix $x$ in $M_n(F)$, think of it as an operator in $L(F^n)$ via the canonical basis of $F^n$. Its trace is then defined canonically as above. And whatever basis you choose for $F^n$, the sum of the diagonal coefficients will be equal to $\mathrm{tr}(x)$. In particular, it is also equal to the sum of the eigenvalues counted with multiplicities when the characteristic polynomial of $x$ splits.
Note It also helps understand why $\mathrm{tr} (ab)=\mathrm{tr}(ba)$, beyond the calculation you mentioned. Indeed
$$
\mathrm{tr}((v_1\otimes w_1^*)(v_2\otimes w_2^*))=w_1^*(v_2)\mathrm{tr}(v_1\otimes w_2^*)=w_1^*(v_2)w_2^*(v_1)
$$
$$
=w_2^*(v_1)w_1^*(v_2)=w_2^*(v_1)\mathrm{tr}(v_2\otimes w_1^*)=\mathrm{tr}((v_2\otimes w_2^*)(v_1\otimes w_1^*))
$$
Best Answer
Whenever you see a matrix trace, you should think inner product, because
$$ \operatorname{Tr}(A^T B) = \langle A, B\rangle_F = \langle A,B\rangle_{\mathbb R^m \otimes \mathbb R^n}$$
that is, the trace of the product of two matrices is equal to their frobenius inner product, which in turn is the induced inner product on the tensor product of Hilbert spaces.
Since it is an inner product, the Cauchy-Schwartz inequality applies:
$$ |\langle A, B \rangle_F |^2 \le \|A\|_F^2\|B\|_F^2$$
with equality if and only if $A$ and $B$ are linearly dependent matrices, i.e. scaler multiples of each other. In your case, we have
$$ |\operatorname{Tr}(ABA^T)| = |\operatorname{Tr}(A^TA B)| = |\langle A^TA , B\rangle_F| \le \|A^TA\|_F\|B\|_F$$
The last term can be further bounded by
$$\begin{aligned} \|A^TA\|_F\|B\|_F &\le \|A\|_F^2\|B\|_F = \Big(\sum\nolimits_i \sigma_i^2(A)\Big)\cdot\sqrt{\sum\nolimits_j \sigma_j^2(B)} \\ &\le rank(A)\cdot \sigma^2_{\max}(A)\cdot rank(B)\cdot\sigma_{\max}(B)\\ &\le m\cdot \min(m,n)\cdot \sigma^2_{\max}(A)\cdot \sigma_{\max}(B) \end{aligned}$$