The blow-up is at most polynomial.

First, given a regular expression $r$, it is straightforward to construct an expression $\tilde r$ using the operators $a\in\Sigma$, $+$, $\cdot$, $(-)^+$, and $\let\nul\varnothing\nul$ such that $$L(\tilde r)=L(r)\let\bez\smallsetminus\bez\{\let\ep\varepsilon\ep\}$$ recursively, by putting $\tilde a=a$, $\tilde\epsilon=\tilde\nul=\nul$, $\let\wt\widetilde\wt{r+s}=\tilde r+\tilde s$, $\wt{r^*}=\tilde r^+$, and $$\wt{r\cdot s}=\tilde r\cdot\tilde s\underbrace{{}+\tilde r}_{\kern-1em\text{if }\ep\in L(s)\kern-1em}\overbrace{{}+\tilde s}^{\kern-1em\text{if }\ep\in L(r)\kern-1em}.$$ Unless $L(r)\subseteq\{\ep\}$, we can subsequently eliminate $\nul$ using $\nul+r=r+\nul=r$, $\nul\cdot r=r\cdot\nul=\nul^+=\nul$. Thus, if $r$ is a regular expression such that $\ep\notin L(r)\ne\nul$, then $\tilde r$ is an $\ep$-free regular expression as defined in the question such that $L(r)=L(\tilde r)$.

In general, the size of $\tilde r$ may be exponential in the size of $r$, but the depth of $\tilde r$ is linear in the depth of $r$. Crucially, every regular expression $r$ of size $n$ has an equivalent regular expression $s$ of depth $O(\log n)$ by Theorem 6.2 in

Moses Ganardi, Markus Lohrey: A universal tree balancing theorem, ACM Transactions on Computation Theory 11 (2019), no. 1, article no. 1, 25 pp, doi 10.1145/3278158. Preprint arXiv:1704.08705 [cs.CC].

Then $\tilde s$ also has depth $O(\log n)$, hence size $2^{O(\log n)}=n^{O(1)}$.

For a fixed alphabet $\Sigma$, the blow-up is at most polynomial.

First, given a regular expression $r$, it is straightforward to construct an expression $\tilde r$ using the operators $a\in\Sigma$, $+$, $\cdot$, $(-)^+$, and $\let\nul\varnothing\nul$ such that $$L(\tilde r)=L(r)\let\bez\smallsetminus\bez\{\let\ep\varepsilon\ep\}$$ recursively, by putting $\tilde a=a$, $\tilde\epsilon=\tilde\nul=\nul$, $\let\wt\widetilde\wt{r+s}=\tilde r+\tilde s$, $\wt{r^*}=\tilde r^+$, and $$\wt{r\cdot s}=\tilde r\cdot\tilde s\underbrace{{}+\tilde r}_{\kern-1em\text{if }\ep\in L(s)\kern-1em}\overbrace{{}+\tilde s}^{\kern-1em\text{if }\ep\in L(r)\kern-1em}.$$ Unless $L(r)\subseteq\{\ep\}$, we can subsequently eliminate $\nul$ using $\nul+r=r+\nul=r$, $\nul\cdot r=r\cdot\nul=\nul^+=\nul$. Thus, if $r$ is a regular expression such that $\ep\notin L(r)\ne\nul$, then $\tilde r$ is an $\ep$-free regular expression as defined in the question such that $L(r)=L(\tilde r)$.

In general, the size of $\tilde r$ may be exponential in the size of $r$, but the depth of $\tilde r$ is linear in the depth of $r$. Crucially, every regular expression $r$ of size $n$ has an equivalent regular expression $s$ of depth $O(\log n)$ by Theorem 6.2 in

Moses Ganardi, Markus Lohrey: A universal tree balancing theorem, ACM Transactions on Computation Theory 11 (2019), no. 1, article no. 1, 25 pp, doi 10.1145/3278158. Preprint arXiv:1704.08705 [cs.CC].

Then $\tilde s$ also has depth $O(\log n)$, hence size $2^{O(\log n)}=n^{O(1)}$.