It's easy to construct continuous surjections onto $(-1,1)^n$ from the $n$-dimensional Euclidean space. See this post, for example: Continuous Surjection from $\mathbb{R}$ onto $\{0,1\}$
However, are there topological obstructions to definition a continuous surjection onto $(-1,1)^n$ with the origin $\{0\}$ removed?
Yes, for $n>1$ there is such surjection. In fact we can generalize this as follows:
Proof. Since each $X_i$ is a Peano space then there is a continuous surjection $\alpha_i:[0,1]\to X_i$ by the Hahn-Mazurkiewicz theorem. Let $D_j=[2\cdot j, 2\cdot j+1]$ and define
$$f:\bigcup_{j=1}^\infty D_j\to X$$ $$f(x)=\alpha_j(x - 2\cdot j)\text{ if }x\in D_j$$
This is a continuous surjection. Now since $X$ is path connected then we can connect each $f(2\cdot j+1)$ with $f(2\cdot j +2)$ by a path $\lambda_j:[2\cdot j+1, 2\cdot j+ 2]\to X$. By glueing all those $\lambda_j$ paths together with $f$ we obtain a continuous surjection $[2,\infty)\to X$ which induces a surjection $\mathbb{R}\to X$ by composing it with $\mathbb{R}\to[2,\infty)$, $x\mapsto x^2+2$. This surjection on the other hand induces $\mathbb{R}^n\to X$ surjections by composing it with any projection $\mathbb{R}^n\to\mathbb{R}$. $\Box$
This easily applies to your case for $X=(-1,1)^m\backslash\{0\}$ when $m>1$. That's because in that case $X$ is path connected and it can be written as a countable union of disks $$D(v_i,r_i)=\big\{v\in\mathbb{R}^m\ \big|\ \lVert v-v_i\rVert\leq r_i\big\}$$ which are Peano spaces. You just take all disks centered at points from $\mathbb{Q}^m$ with rational radii and exclude those that are not fully contained in $X$. This works for any open subset of $\mathbb{R}^m$. And indeed, any open subset $U$ of any $\mathbb{R}^m$ is an image of any $\mathbb{R}^n$ as long as $U$ is path connected.
Note that for $m=1$ $X$ is not an image of any $\mathbb{R}^n$ since it is not path connected.