Suppose we have a simple Lie group $G$ with algebra $\mathfrak{g}=\{X_a\}$, where the generators $X_a$ are in some matrix representation. Is it true that the only invariant rank $n$ tensor in the adjoint representation is
$$ \hspace{80pt} T_{a_1\cdots a_n}={\rm Tr} (X_{a_1}\cdots X_{a_n}) \ ? \hspace{80pt} (*) $$
Hint:
$T$ is clearly invariant, under $g\in G$ it transforms as
$$ \begin{split} T_{a_1\cdots a_n}\xrightarrow{\ g\ } & \ \sum_{b_1\ldots b_n}D_{\rm adj}(g)_{a_1}{}^{b_1}\cdots D_{\rm adj}(g)_{a_n}{}^{b_n}T_{b_1\cdots b_n} \\ &={\rm Tr}(g^{-1}X_{a_1}g\cdots g^{-1}X_{a_n}g)\\ &= T_{a_1\cdots a_n}\, . \end{split} $$ ($D_{\rm adj}(g)$ is the group element $g$ in the the adjoint representation, and in the second line $g$ is in the same matrix representation as the generators.) However I do not know whether all invariant tensors can be written in the form $(*)$.
Example:
The statement $(*)$ is true for $G=SO(3)$, where it is well known that the only tensors invariant under rotations are $\delta_{ij}$ and $\epsilon_{ijk}$ and tensor products of these two (which allows one to build the rotational-invariant scalar and cross products in $\mathbb{R}^3$). Using a matrix representation of the algebra (e.g. Pauli matrices), it is easy to see that
$$ \delta_{ij} \propto {\rm Tr} (\sigma_i\sigma_j) \hspace{20pt}\hbox{and}\hspace{20pt} \epsilon_{ijk}\propto {\rm Tr} (\sigma_i\sigma_j\sigma_k)\, , $$ showing that tensors of the form $(*)$ exhaust all possible invariant tensors.
This is a somewhat incomplete answer, but it could be generalised somehow. It works at least for SU(n) and SO(n). An invariant tensor is a scalar under a group transformation, and since it is a tensor of rank $n$ it must be built by tensoring up $n$ adjoint representations $Ad \otimes \cdots \otimes Ad$, the fact is that of all the tensors that arise in the decomposition of this tensor product into irreducible representations, the invariant tensors are those wich correspond to the ${\bf 1}$ representation, that is we have:
\begin{equation} Ad \otimes \cdots \otimes Ad = \nu_n {\bf 1} \oplus \cdots \end{equation}
This means that the components of the invariant tensor are essentially the Clebsch-Gordan coefficients coupling the l.h.s. to the trivial representation in the r.h.s. Now, these Clebsch-Gordan coefficients must couple the unit vectors to a scalar, and for instance in the example you gave $(\sigma_i)^{\alpha}_{\beta}$ can be thought of as the Clebsch-Gordan coupling one fundamental in $\beta$ and one anti-fundamental (which are the same for SU(2)) in $\alpha$ to a vector in $i$. That is to say that the Pauli matrices are, in some base, just these Clebsch-Gordan coefficients. I suspect a similar argument can be made for the $(X_a)^b_c$, and if this is so, then it follows that the only one way to couple all these matrix elements to a scalar is to trace them into some combination. This means in particular that the answer to the question is no. You can form invariant tensors that are tensor products of two smaller rank invariant tensors, for instance take $n \geq 6$:
\begin{equation} T_{a_1 \cdots a_n} = Tr(X_{a_1}X_{a_2})Tr(X_{a_3}X_{a_4}X_{a_5})Tr(X_{a_6} \cdots X_{a_n}) \end{equation}
First, it can be seen that the $(X_a)^{b}_{c}$ are indeed Clebsch-Gordan coefficents, for they project a product of two vectors in $Ad \otimes Ad = {\bf 1}\oplus Ad \oplus \cdots $ to the vector in $Ad$ on the l.h.s, componentwise they take $u^a,v^b$ to $w^a=f^{a}_{bc}u^b v^c$. Then, look at the product of two adjoint representations
\begin{equation} Ad \otimes Ad = {\bf 1} \oplus Ad \oplus D := d_2 \end{equation} Where the ${\bf 1}$ corresponds to $\phi= \delta_{ab} u^{a}v^{b}$ and $Ad$ to $w_a$ above. ($\phi$ is scalar when $Ad$ is orthogonal, which happens whenever $G$ is compact) Also notice that $D$ is self-conjugate so $\bar{D} = D$. Now consider the product of $4$ adjoint representations
\begin{equation} ({\bf 1} \oplus Ad \oplus D) \otimes ({\bf 1} \oplus Ad \oplus D) = {\bf 1}\oplus {\bf 1} \oplus {\bf 1} \oplus Ad \oplus Ad \oplus Ad \oplus \sum_{i=1}^3 D^{(1)}_{i} := d_{4} \end{equation} where all $D^{(1)}_i$ are self-cojugate also. Observe that since $D$ is made of two antifundamental and two fundamental irreps, the sum runs $i=1,2,3$. It is now a matter of combinatorics to compute the degeneracy of ${\bf 1}$, $\nu_n$ above. This number should be equal to the number of independent ways to trace $n$ generators in the adjoint representation, this can in principle be done recursively, although the details are somewhat messy.
We can give a combinatoric expression for this. Notice that in order to trace we must trace two or more generators (there's no such thing as an invariant tensor of rank 1), hence we must look at the partitions of $n=k_1 + k_2 + \cdots + k_r$ with each $k_i \geq 2$, there are $\binom{n-r-1}{r-1}$ such partitions, and for each one we may put the indices in $n!$ ways, however we must account for cyclicity of the trace and hence divide by $k_1 k_2 \cdots k_r$, me must also divide by $r!$ because the ordering of different traces is unimportant. The total number of rank $n$ invariant tensors is now given by the sum below, which should equal $\nu_n$
\begin{equation} \sum_{r=1}^{\lceil n/2 \rceil} \binom{n-r-1}{r-1} \sum_{ \sum_{i}^{r}k_i =n,~k_i \geq 2} \frac{n!}{r! k_1 \cdots k_r} = \nu_n \end{equation} To show this however, is a separate question, and is a problem of combinatorics.