Claim: The minimum $1/e \approx 0.36788$ is achieved at $x = e^{-e} \approx 0.065988$.
We proceed by first establishing that the minimum is upper bounded by $1/e$ and then show that it is also lower bounded by $1/e$.
Proof of upper bound. Let $y_{0}=1$ and define the iteration $$ y_{n+1}=x^{y_{n}}. $$ Note that $$ y_{1}=x,\qquad y_{2}=x^{x},\qquad\text{etc.} $$ tends to the infinite tetration. It is known that this iteration converges (possibly to $\infty$) for all $x\geq e^{-e}$. Define $y_{\infty}$ as the function which associates to each $x \geq e^{-e}$ the limit $\lim_{n\rightarrow\infty}y_{n}$. Note that if $y_\infty(x)$ is finite, then $$ y_{\infty}(x)=\lim_{n\rightarrow\infty}y_{n+1}=\lim_{n\rightarrow\infty}x^{y_{n}}=y^{y_{\infty}(x)} $$ and hence $$ (y_{\infty}(x))^{1/y_{\infty}(x)}=x. $$ Solving for $y_{\infty}(x)$, $$ y_{\infty}(x)=\frac{W(\log(1/x))}{\log(1/x)}. $$ Let $x_0=1$ and define also the iteration $$ x_{n+1}=x^{x^{x_{n}}}. $$ Since it is monotone, this iteration converges (possibly to $\infty$) for all $x\geq0$. Define $x_{\infty}$ as the function which associates to each $x \geq 0$ the limit $\lim_{n\rightarrow\infty}x_{n}$. Since $x_{\infty}=y_{\infty}$ on $[e^{-e},\infty)$, $$ \min_{x\geq0}x_{\infty}(x)\leq y_{\infty}(e^{-e})=1/e. $$
Proof of lower bound. Fix $x$ and let $a=x_{\infty}(x)$ for brevity. Then, $$ x^{x^{a}}=a\implies x=a^{1/(x^{a})}\implies x^{a}=a^{a/(x^{a})}\implies(x^{a})^{x^{a}}=a^{a}. $$ Therefore, $$ a^{a}\geq\min_{u\geq0}u^{u}=e^{-1/e}, $$ which in turn implies $ a\geq1/e $.