$\newcommand{\bs}[1]{{\boldsymbol #1}}
\newcommand{\tildebssigma}{\tilde{\bs\sigma}}
\newcommand{\bssigma}{{\bs\sigma}}$Yes, products of Pauli matrices form a basis for the set of Hermitian matrices (of dimensions that are powers of $2$).
More specifically, fix an integer $n$ and let $N\equiv 2^n$, define $\bssigma\equiv(\sigma_x,\sigma_y,\sigma_y)$, and $\tildebssigma\equiv (I,\sigma_x,\sigma_y,\sigma_y)$.
Moreover, let $J\in\{0,1,2,3\}^{n}$ be a tuple of $n$ integers with each $J_i\in\{0,1,2,3\}$.
Consider the matrices of the form
$$
\tildebssigma_J\equiv\prod_{k=1}^n \tildebssigma^{(k)}_{J_k},
$$
where $\tildebssigma_j^{(k)}$ denotes the Pauli matrix $\tildebssigma_j$ applied on the $k$-th qubit.
You can check that these are all Hermitians. Moreover, $\tildebssigma_J$ is traceless for all $J\neq(0,...,0)$ (for which $\tildebssigma_{(0,...,0)}=I$).
There are $4^n=N^2$ matrices of this form (one for each possible choice of $J$), and for any pair of tuples $J,K$ we have
$\operatorname{Tr}(\tildebssigma_J\tildebssigma_K)=N\delta_{JK}$.
Moreover, the space of Hermitian $N\times N$ matrices also has dimension $N^2$. It follows that $\{\tildebssigma_J\}_J$ is a basis for this space. Explicitly, you can decompose an arbitrary Hermitian matrix $H$ as
$$H=\frac{1}{N}\sum_J \operatorname{Tr}(\tildebssigma_J H)\tildebssigma_J.$$
Note that the coefficients in any such expansion are always real. This is not by chance. Indeed, being more careful, we should state that the set of real linear combinations of products of Pauli matrices give the set of Hermitian matrices (notice that if $A$ is Hermitian then $\alpha A$ is Hermitian iff $\alpha\in\mathbb R$, so this is not surprising).
More general matrices can be generated if we allow for complex coefficients in the expansion.
Indeed, note that
$$\frac{I+Z}{2}=\begin{pmatrix}1&0 \\ 0 & 0\end{pmatrix},
\qquad \frac{I-Z}{2}=\begin{pmatrix}0&0 \\ 0 & 1\end{pmatrix}, \\
\frac{X+iY}{2}=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}
\qquad
\frac{X-iY}{2}=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}.
$$
Therefore the complex span of Pauli matrices can be used to generate arbitrary $2\times 2$ matrices. This then translates into the same result for arbitrary $2^n$-dimensional spaces, as if $\mathcal V\equiv\{v_k\}$ is a basis for $V$, then the sets of tensor products of elements of $\mathcal V$ form a basis for $V^{\otimes n}$.
This is a special instance of the more general fact that any matrix can be decomposed uniquely as sum of a Hermitian and a skew-Hermitian matrix, as
$$A=\frac{A+A^\dagger}{2}+i\frac{A-A^\dagger}{2i},$$
and the fact that products of Pauli matrices give you a basis for the set of Hermitian matrices.