On the Wikipedia page one can asympotoic formula of the Bessel function
$$ I_0(z) \propto \frac{e^z}{\sqrt{2\pi z}} $$
On the Wolfram page there is a more detailed asymptotic formula for the Bessel functions
$$ I_0(z) \propto \frac{1}{\sqrt{2\pi z}} \left[ e^z \,{}_2F_0(\tfrac{1}{2}, \tfrac{1}{2};; \tfrac{1}{2z})+ i \, e^{-z}\,{}_2F_0(\tfrac{1}{2}, \tfrac{1}{2};; -\tfrac{1}{2z}) \right] $$
I don't know much about hypergeometric functions except for this identity:
$$ {}_2F_1(a,b,c; z) = \int_0^1 x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a} \, dx$$
Is it possible to re-write the above asymptotic in terms of Gauss hypergeometric functions instead of generalized hypergeometric functions?
It is not possible to rewrite the above asymptotic in terms of Gauss hypergeometric function $_2F_1$ because it is a degenerative form : The series representation of $_2F_0(a,b,x)$ is a divergent hypergeometric series. However, for $x < 0$ we have $$_2F_0(a,b,x) = \left(\frac{-1}{x}\right)^a U(a,1+a-b,-1/x)$$ where $U$ is the symbol for the confluent hypergeometric function, also known as Tricomi function. http://mathworld.wolfram.com/ConfluentHypergeometricFunctionoftheSecondKind.html
So, you can rewrite the asymptotic series in terms of confluent hypergeometric functions: $$_2F_0\left(\frac{1}{2} ,\frac{1}{2}, \frac{1}{2z}\right) = (-2z)^{\frac{1}{2}}\: U\left(\frac{1}{2},1,-2z\right)$$