First, we can observe that your integral depends solely on the behavior of $f$ on the interval $[0, 1]$. Its values outside that region do not affect the expression. So we may multiply by the cutoff function $\chi_{[0, 1]}$.
Thus we may consider this problem for elements of $L^2(\mathbb{R})$ with compact support.
We can look at the Hardy-Littlewood-Sobolev theorem on fractional integration (or more accurately, its proof). The argument found in Remark 2 here, partitioning into dyadic shells, shows that for $x > 0$, $$\left|\int_0^x \frac{f(y)}{|x - y|^{1/2}} \, dy \right| \leq \int_0^x \frac{|f(y)|}{|x - y|^{1/2}} \, dy \leq \int_{B(x, x)} \frac{|f(y)|}{|x - y|^{1/2}} \, dy \leq C x^{1/2} M f(x),$$ where $M f$ denotes the Hardy-Littlewood maximal function and $C$ is an absolute constant.
In your case, since we are integrating over $x \in [\epsilon, 1]$, we can bound this solely by a multiple of $M f$, and then the strong-type
Hardy-Littlewood $L^p$ estimate gives (as a very rough upper bound), $$\left\|\int_0^x f(y) |x - y|^{-1/2} \, dy \right\|_{L^2_x[0, 1]} \leq C \|M f(x)\|_{L^2_x} \leq C' \|f\|_{L^2} = C' \|f\|_{L^2[0, 1]},$$ invoking the support restriction on $f$.
So your integral is bounded above by the quantity $K \int_{0}^{1} |f(x)|^2 \, dx$, for some dimensional constant $K$. (And this also indicates that you don't need to take $\epsilon > 0$; you can directly integrate over $[0, 1]$.) You specified $f$ is locally $L^2$, so this quantity is well-defined and non-infinite.
Thus, for any $f \in L^2_{\text{loc}}$, you can guarantee that your integral will be finite.
$\newcommand\al\alpha\newcommand\EE{\mathcal E}\newcommand\ip[2]{\langle #1,#2\rangle}$The answer is
$$\mu(\EE(S^1))=\mu(H^\al(S^1))=0$$
for all real $\al\ge0$.
Indeed, since $\EE(S^1))\subseteq H^\al(S^1)\subseteq H^0(S^1)=L^2(S^1)=:L^2$
for all real $\al\ge0$, it is enough to show that $\mu(L^2)=0$.
Let $X$ be a random vector in $\EE'(S^1)$ with distribution $\mu$. Let $(e_1,e_2,\dots)$ be an orthonormal basis of $L^2$, with $e_n\in\EE(S^1)$ for each $n$. Then $X_1:=X(e_1),X_2:=X(e_2),\dots$ are independent standard normal random variables and hence on the event $X\in L^2$ we have $\|X\|_{L^2}^2=\sum_{n=1}^\infty X_n^2=\infty$ almost surely (a.s.) -- say, by the strong law of large numbers.
So, $P(X\in L^2)=0$; that is, $\mu(L^2)=0$. $\quad\Box$
Details on the latter two sentences: By the strong law of large numbers, $$\sum_{n=1}^\infty X_n^2
=\lim_{N\to\infty}N\frac1N\,\sum_{n=1}^N X_n^2 \\
=\lim_{N\to\infty}N \; \lim_{N\to\infty}\frac1N\,\sum_{n=1}^N X_n^2
=\lim_{N\to\infty}N \times 1=\infty\text{ a.s.}$$
Introducing now the events $A:=\{X\in L^2\}$ and $B:=\{\sum_{n=1}^\infty X_n^2=\infty\}$, we see that $A\cap B=\emptyset$ and $P(B)=1$. So,
$$P(X\in L^2)=P(A) \\
=P(A\cap B)+P(A\setminus B)=0+0=0.$$
Best Answer
The maximum $x_n$ of $$f_n(x):=e^{-1/x}\Bigl(1+\frac{1}{n^2 x^n} \Bigr)$$ is the smallest solution in $(0,1)$ of the equation $$x=n x^n+\frac{1}{n}.$$ For $n\gg 1$ this gives $x_n\rightarrow 1/n$.
The integral is given by $$\int_0^1 f_n(x)dx=\text{Ei}(-1)+n^{-2}\,\Gamma (n-1,1)+1/e$$ $$\qquad\rightarrow \sqrt{2 \pi } e^{-n} n^{n-\frac{7}{2}}\;\;\text{for}\;\;n\gg 1.$$
Here is a comparison of the exact integral (gold data points) and the asymptote (blue) --- the difference is hardly noticeable for $n>10$.