The fastest way I know (admittedly a bit sloppy) to come to solutions is:
- From symmetry we know there is only one vector potential component $A_\varphi$, assuming cylindrical coordinates and having the loop in the $z=0$ plane.
- If we look outside the wire we can treat the current as if it flows only on the center-line (even though it is on the surface, or distributed, the same trick as treating the mass of the Earth as being in the center).
Let $R$ be the radius of the loop and let's integrate over the center-line current with the Green function to find $A_\varphi$, which will only depend on $\rho$ and $z$:
$$\begin{align}
A_\varphi(\rho,z) =&\ \frac{\mu_0\ I}{4\pi}\int\limits_0^{2\pi}\!\!R\,d\varphi\
\frac{\cos\varphi}{\sqrt{(R-\rho \cos\varphi)^2+(\rho \sin\varphi)^2+z^2}}
\\[5pt]
=&\ \frac{\mu_0\ I\ R}{4\pi}\int\limits_0^{2\pi}\!\!d\varphi\
\frac{\cos\varphi}{\sqrt{R^2+\rho^2+z^2-2R\rho \cos\varphi}}
\\[5pt]
=&\ \frac{\mu_0\ I\ R}{4\pi\sqrt{R^2+\rho^2+z^2}}\ \int\limits_0^{2\pi}\!\!d\varphi\
\frac{\cos\varphi}{\sqrt{1-u\, \cos\varphi}}
\\[5pt]
=&\ \frac{\mu_0\ I\ R}{4\pi\sqrt{R^2+\rho^2+z^2}}\
\Big[\frac{-4\sqrt{1+u}}{u}\ E\big(\frac{2u}{1+u}\big)
+\frac{4}{u\,\sqrt{1+u}}\ K\big(\frac{2u}{1+u}\big) \Big]
\end{align}$$
where after defining $u=2 R \rho/(R^2+\rho^2+z^2)$, the integral was found with Mathematica. This can further be reduced to:
$$\begin{align}
A_\varphi(\rho,z) =&\ \frac{\mu_0\, I\,\sqrt{(R+\rho)^2+z^2}}{2\pi\,\rho}
\ \Big[-E(m) +\big(1-\frac m2 \big)\ K(m) \Big]
\\[6pt]
&\text{with}\ \ m=\frac{4R\,\rho}{(R+\rho)^2+z^2}
\end{align}$$
In these expressions the complete Elliptic integrals $E(m)$ and $K(m)$ are defined as in Mathematica (see [question 4901060] for the definition ambiguity).
For the magnetic field, ${\bf B}={\bf \nabla}\times{\bf A}$, we use the expressions in cylindrical coordinates from [Jackson]. This leads to:
$$\begin{align}
{\bf B}_\rho(\rho,z) =&\ \frac{\mu_0\,I\, z\ \left[(r^2+R^2+z^2)\,E(m) -\left((r-R)^2+z^2\right)K(m) \right]}{2 \pi\, r\,
\big((r-R)^2+z^2\big) \sqrt{(r+R)^2+z^2}}
\\[8pt]
{\bf B}_z(\rho,z) =&\ \frac{\mu_0\,I\ \left[\big(r^2 (z^2-R^2)+(R^2+z^2)^2\big)\,E(m) - (R^2+z^2) \big((r-R)^2+z^2\big) K(m) \right]}{2 \pi\, r^2\,
\big((r-R)^2+z^2\big) \sqrt{(r+R)^2+z^2}}
\end{align} $$
If $r_w$ is the wire diameter, then the field right at the surface of the wire is found as ${\bf B}_\rho(R,r_w)$, the horizontal field just above the wire at heigth $z=r_w$, or also as ${\bf B}_z(R-r_w,0)$, the vertical field at the inner side.
So we ask Mathematica for some limits close to the current loop :
Limit[ z Br /.r->R, z->0, Direction->"FromAbove", Assumptions->R>0]
Limit[ w Bz /.{r->R-w,z->0}, w->0, Direction->"FromAbove", Assumptions->R>0]
and in both cases get mu0 I/(2Pi). So this proves that, as expected:
$$\begin{align}
{\bf B}_\rho(R,r_w) \sim &\ \frac{\mu_0\,I}{2 \pi\, r_w}
\\[6pt]
{\bf B}_z(R-r_w,0) \sim &\ \frac{\mu_0\,I}{2 \pi\, r_w}
\end{align} $$
so the field simply goes inversely proportional to the distance from the wire, the same as for a straight wire. Other limits we can look at:
Limit[Bz, r->0]//Simplify
Limit[Bz, z->0]//Simplify
The first one is the field on the $z$-axis that was already meantioned in the question, the second is the field in the plane of the loop. Results:
$$\begin{align}
{\bf B}_z(0,z) =&\ \frac{\mu_0\,I\, R^2}
{4\, (R^2+z^2)^{3/2}}
\\[8pt]
{\bf B}_z(\rho,0) =&\ \frac{\mu_0\,I\ R^2}{2 \pi\, \rho^2\,
(R^2-\rho^2)}
\Big[\ (\rho+R)\ E\Big(\frac{4\,\rho\,R}{(\rho+R)^2}\Big)
+(\rho-R)\ K\Big(\frac{4\,\rho\,R}{(\rho+R)^2}\Big) \Big]
\end{align} $$
If this last one is plotted for $R=1, I=1$, it looks like this:
