We can show more generally that if $\mathcal M$ is a linear subspace of $\mathcal M_n(\mathbb R)$ such that all its element have a rank less than or equal to $p$, where $1\leq p<n$, then the dimension of $\mathcal M$ is less than or equal to $np$. To see that, consider the subspace $\mathcal E:=\left\{\begin{pmatrix}0&B\\^tB&A\end{pmatrix}, A\in\mathcal M_{n-p}(\mathbb R),B\in\mathcal M_{p,n-p}(\mathbb R) \right\}$. Its dimension is $p(n-p)+(n-p)^2=(n-p)(p+n-p)=n(n-p)$. Let $\mathcal M$ a linear subspace of $\mathcal M_n(\mathbb R)$ such that $\displaystyle\max_{M\in\mathcal M}\operatorname{rank}(M)=p$. We can assume that this space contains the matrix $J:=\begin{pmatrix}I_p&0\\0&0\end{pmatrix}\in\mathcal M_n(\mathbb R)$. Indeed, if $M_0\in\mathcal M$ is such that $\operatorname{rank}M_0=p$, we can find $P,Q\in\mathcal M_n(\mathbb R)$ invertible matrices such that $J=PM_0Q$, and the map $\varphi\colon \mathcal M\to\varphi(\mathcal M)$ defined by $\varphi(M)=PMQ$ is a rank-preserving bijective linear map.
If we take $M\in\mathcal M\cap \mathcal E$, then we can show, considering $M+\lambda J\in\mathcal M$, that $M=0$. Therefore, since
$$\dim (\mathcal M+\mathcal E)=\dim(\mathcal M)+\dim(\mathcal E)\leq \dim(\mathcal M_n(\mathbb R))=n^2, $$
we have
$$\dim (\mathcal M)\leq n^2-n(n-p)=np.$$
The dimensions you are obtaining for $V$, the $\mathbb R$-vector space of symmetric matrices, and $W$, the subspace of traceless symmetric matrices, are not correct. So your arguments cannot be completely valid, sorry.
And yes, there are fundamental mistakes.
It appears (to me) as if you are still struggling with the notion of a basis.
Several (equivalent) definitions are available, e.g., the following one which suits the present context:
A system of vectors $b_1, b_2,\dots , b_d\in X$ is called a basis for the vector space $X$ if any vector $x\in X$ admits a unique representation as a linear combination
$$\alpha_1b_1+\alpha_2b_2+\ldots +\alpha_db_d\,=\,x\,.$$
That a basis always exists, and that different bases of the same vector space
have equal cardinality are two (proven) fundamental consequences. On the one hand they allow for attributing the dimension $d$ directly to the vector space, without referring to a chosen basis. On the other hand, any suitable basis can be considered, to read off the dimension of the vector space.
If $M_n(\mathbb R)$ denotes the $\mathbb R$-vector space of real $n\times n$ matrices, let $E_{ij}\in M_n(\mathbb R)$ be the matrix equal to $1$ at position $i,j$ and zero elsewhere. The $E_{ij}$ with $i,j=1,\dots,n\,$ constitute a basis of $M_n(\mathbb R)$, the uniqueness according to the above definition is fairly obvious. And $\,\dim M_n(\mathbb R)=n^2$.
Symmetry of a matrix $A$ means $a_{ji}=a_{ij}$ for all its entries (for the diagonal elements this condition is void), which is a restriction compared to $M_n(\mathbb R)$ as the upper triangular matrix entries of $A$ fix the entries in the strictly lower triangular corner. This leads to considering the system
$E_{ij}+E_{ji}$ where $1\leqslant i<j\leqslant n$, joint with $E_{ii}, 1\leqslant i\leqslant n$, as a candidate for a basis of $V$. Again, checking the required uniqueness is straightforward, so this is indeed a basis of the symmetric matrices. As a result
$$\dim V \:=\:\frac{(n-1)\,n}{2}+n \:=\: \frac{n\,(n+1)}{2}$$
Imposing $\,\operatorname{tr}A=0$ is an additional constraint on elements of $V$, hence $\dim W<\dim V$ is expected: Invoking Rank-Nullity as in the OP yields
$$\dim W \:=\:\dim V -\dim\operatorname{Im}(\operatorname{tr}: V\to\mathbb R)
\:=\:\frac{n\,(n+1)}{2}-1 \:=\:\frac{(n-1)(n+2)}{2}$$
Alternatively, you may take $V$'s basis as of above and subtract from each element its trace times $E_{nn}$. This leaves the $E_{ij}+E_{ji}$ unchanged, and gives $F_{ii}:=E_{ii}-E_{nn}$ where $1\leqslant i\leqslant n-1$, hence at this stage the dimension goes down by one. (Generating sets may contain the null vector $F_{nn}=0$, whereas a basis cannot, never ever.)
Checking uniqueness and the "final count down" is left to you.
Hope this reply is helpful to you.
When searching for the terms "dim symmetric trace" I found the site-internal references (being so closely related that your post probably gets marked as duplicate)
Best Answer
HINT
If the columns of $A$ are $A_1, A_2, A_3$ your constraint says $$ 1A_1 +2A_2 + 3A_3 = 0 \iff A_1 = -2A_2 - 3A_3, $$ which means that your matrix looks like $$ A = \left[ -2A_2-3A_3,A_2,A_3 \right], $$ so how many free parameters do you really have?