A homological obstruction that I mentioned in the comments can be extracted from On homotopical and homological $Z_n$-sets by Taras Banakh, Robert Cauty, and Alex Karassev.

This is a long paper but we won't need to know much. Given an abelian group $G$, a point $x\in X$ is a $G$-homological $Z_n$-set if $H_k(X, X-\{x\}; G)=0$ for all $k\le n$. The definition reduces to the one given on page 1 because $H_k(X, X-\{x\}; G)=H_k(U, U-\{x\}; G)$ for any open neighborhood $U$ of $x$ by excision in homology. For example, $0\in [-1,1]$ is a $Z_0$-set but not a $Z_1$-set.

Now suppose $G$ is $\mathbb Q$ or $\mathbb Z_p$, and $X$ is a Tychonov space. Theorem 6.1 implies that if a point $x\in X$ is a $G$-homological $Z_n$-set, then the point $(x,0)$ is a $G$-homological $Z_{n+1}$-set in $X\times [-1,1]$. Therefore, if a neighborhood of $x\in X$ is homeomorphic to a neighborhood of $(x,0)\in X\times [-1,1]$, then $n=n+1$, and hence $n=\infty$. The latter is impossible if $x\in X$ is a $Z_n$-set but not a $Z_{n+1}$-set (over $G$), which gives an obstruction.

Clearly, no obstruction occurs when $X$ is the product of infinitely many copies of $\mathbb R$.

A homological obstruction that I mentioned in the comments can be extracted from On homotopical and homological $Z_n$-sets by Taras Banakh, Robert Cauty, and Alex Karassev.

This is a long paper but we won't need to know much. Given an abelian group $G$, a point $x\in X$ is a $G$-homological $Z_n$-set if $H_k(X, X-\{x\}; G)=0$ for all $k\le n$. The definition reduces to the one given on page 1 because $H_k(X, X-\{x\}; G)=H_k(U, U-\{x\}; G)$ for any open neighborhood $U$ of $x$ by excision in homology. For example, $0\in \mathbb R$ is a $Z_0$-set but not a $Z_1$-set.

Now suppose $G$ is $\mathbb Q$ or $\mathbb Z_p$, and $X$ is a Tychonov space. Theorem 6.1 implies that if a point $x\in X$ is a $G$-homological $Z_n$-set, then the point $(x,0)$ is a $G$-homological $Z_{n+1}$-set in $X\times \mathbb R$.

Therefore, if every point of $X$ is a $Z_n$-set, and a neighborhood of some $x\in X$ is homeomorphic to a neighborhood of $(y,t)\in X\times \mathbb R$, then $n=n+1$, and hence $n=\infty$. The latter is impossible if $x\in X$ is not a $Z_{n+1}$-set, which gives an obstruction.

Note that no obstruction occurs when $X$ is the product of infinitely many copies of $\mathbb R$, and in fact, in this case $X$ is homeomorphic to $X\times \mathbb R$.

To apply the above one only needs to check whether the local homology is trivial (over $\mathbb Q$ or $\mathbb Z_p$). Computing the local homology is not necessary.