$\newcommand{\bs}[1]{{\boldsymbol #1}}
\newcommand{\tildebssigma}{\tilde{\bs\sigma}}
\newcommand{\bssigma}{{\bs\sigma}}$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.

I realised after answering that this is essentially a duplicate of https://quantumcomputing.stackexchange.com/q/2703/55

– glS – 2019-11-10T00:22:01.503The statement $\mathfrak{su}(n)_{\mathbb C}=\mathfrak{gl}(n,\mathbb C)$ is not true. Out of curiosity, where did you see this? – Jonathan Trousdale – 2019-11-11T04:53:28.303

@ChainedSymmetry right, it's just me always messing up the details. It's clearly wrong because complexifying $\mathfrak{su}$ we always get traceless matrices. It should be right that $\mathfrak{u}(n)_{\mathbb C}=\mathfrak{gl}(n,\mathbb C)$ though, no? See e.g. https://math.stackexchange.com/q/2289677/173147.

– glS – 2019-11-11T09:14:29.550@ChainedSymmetry anyway, I removed the comment about lie algebra, it was not that relevant anyway for the decomposition I was thinking of – glS – 2019-11-11T09:57:57.553

That's correct, as long as you mean "isomorphic to" by "$=$" (it's more common to use "$\cong$" to avoid confusion). – Jonathan Trousdale – 2019-11-11T14:32:19.957