Global Minimum of $f(a) = \int _{-\infty}^{\infty} \exp\left(-|x|^a\right)dx, a\in(0,\infty)$

$$\int_0^{\infty} dx \, e^{-x^a} = \frac1{a} \int_0^{\infty} du \, u^{1/a - 1} e^{-u} = \frac1{a} \Gamma \left ( \frac1{a} \right ) = \Gamma \left ( 1+\frac1{a} \right )$$

To find a local minima, use $\Gamma'(x) = \Gamma(x) \psi(x) $. Other than that, you are more or less stuck with a transcendental equation. Certainly, $\psi(x)$ has a zero in the vicinity of the minimum you have pointed out, but I do not have a good, non-numerical way of locating it.

Continuing from Ron Gordon's answer, we just have to locate the absolute minimum of the function $\Gamma(1+z)$ over $\mathbb{R}^+$. Since $\Gamma(1)=\Gamma(2)=1$, it is trivially in the interval $I=(0,1)$. Since: $$\Gamma'(1+z) = \Gamma(1+z)\,\psi(z+1)\tag{1}$$ we just have to find the only solution in $I$ of: $$ \gamma = \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right).\tag{2}$$ The RHS is a smooth, increasing, concave function over $I$, so just a few steps of Newton's method with starting point $z=\frac{1}{2}$ give that the solution is about $0.461632$, so the minimum of $\Gamma(z+1)$ over $\mathbb{R}^+$ is about $\color{red}{0.8856}$.

Continuing from Ron Gordon's answer, $$\frac {d}{da}\Gamma(1+\frac 1a)=-\frac{\Gamma \left(1+\frac{1}{a}\right) \psi \left(1+\frac{1}{a}\right)}{a^2}$$ and so, we look for the zero of $\psi \left(1+\frac{1}{a}\right)$.

Taking into account your result, we can expand $\psi \left(1+\frac{1}{a}\right)$ as a Taylor series built at $a=2$ and obtain $$\psi \left(1+\frac{1}{a}\right)=\psi \left(\frac{3}{2}\right)+\left(1-\frac{\pi ^2}{8}\right) (a-2)+\frac{1}{32} (a-2)^2 \left(-16+2 \pi ^2+\psi ^{(2)}\left(\frac{3}{2}\right)\right)+\frac{1}{384} (a-2)^3 \left(192-12 \pi ^2-\pi ^4-12 \psi ^{(2)}\left(\frac{3}{2}\right)\right)+O\left((a-2)^4\right)$$ Numerically $$\psi \left(\frac{3}{2}\right)\approx 0.036489973978576520559$$ $$\psi ^{(2)}\left(\frac{3}{2}\right)\approx -0.82879664423431999560$$ Limiting to the second order, solving the quadratic equation leads to $$a \approx 2.16699$$ Using the third order would lead to $$a \approx 2.16618$$ quite close to your observation which is the perfect result.