It looks like your question is really about finding the joint distribution of $X_{1:n}$ and $\overline{X}$, so that's the one I'm going to address. We'll do this by finding the marginal distribution of $X_{1:n}$ and the conditional distribution of $\overline{X}$ given $X_{1:n}$. Multiplying them together will give the joint distribution.
First, the marginal of $X_{1:n}$. The distribution of the minimum $X_{1:n}$ of $n$ independent exponential random variables is again exponential, and it's not too hard to prove that a similar result holds for the exp$(\theta, \eta)$ distribution. We have that $X_{1:n}$ is exp$(\theta/n,\eta)$; i.e., has pdf $f_{X_{1:n}}(x) = \frac{n}{\theta} e^{n(\eta - x)/\theta} {\bf 1}_{(x \geq \eta)}.$
Then, the conditional distribution of $\overline{X}$ given $X_{1:n}$. Given that $X_{1:n} = a$, each of the other $n-1$ variables from the sample has the cdf $P(X < x| X \geq a)$, where $X$ is exp$(\theta, \eta)$. As with the memoryless property of the usual exponential distribution, it is not too hard to show that each of these other variables is exp$(\theta,a)$; i.e., has pdf $\frac{1}{\theta} e^{(a-x)/\theta} {\bf 1}_{(x \geq a)}.$
The distribution of $\overline{X}$ given $X_{1:n} = a$, then, is the distribution of $\frac{1}{n}(X_1 + X_2 + \cdots + X_{n-1} + a)$, where $X_i$ is exp$(\theta,a)$. The moment generating function of $\overline{X}$ given $X_{1:n} = a$ can then be found (using standard mgf techniques) to be $\frac{e^{at}}{(1- \theta t/n)^{n-1}}.$
Using the time shift property of the Laplace transform (or by direct calculation) we can see that this is the mgf of what we might reasonably call a gamma$(n-1,\theta/n,a)$ distribution. (Since the mean of exponentials is gamma, it is reasonable that the mean of shifted exponentials is a shifted gamma.) Thus the pdf of $\overline{X}$ given $X_{1:n} = a$ is $\left(\frac{n}{\theta}\right)^{n-1} \frac{1}{\Gamma(n-1)} (\bar{x} - a)^{n-2} e^{n(a - \bar{x})/\theta} {\bf 1}_{(\bar{x} \geq a)}.$
Finally, compute the joint distribution. Since $f_{X_{1:n},\overline{X}}(a,\bar{x}) = f_{X_{1:n}}(a) f_{\overline{X}|a}(\bar{x}|a)$ we have that
$f_{X_{1:n},\overline{X}}(a,\bar{x}) = \frac{n}{\theta} e^{n(\eta - a)/\theta} \left(\frac{n}{\theta}\right)^{n-1} \frac{1}{\Gamma(n-1)} (\bar{x} - a)^{n-2} e^{n(a - \bar{x})/\theta} {\bf 1}_{(\bar{x} \geq a)} {\bf 1}_{(a \geq \eta)}.$ Simplifying, the joint pdf of $X_{1:n}$ and $\overline{X}$ that you're looking for is $f_{X_{1:n},\overline{X}}(a,\bar{x}) = \left(\frac{n}{\theta}\right)^n \frac{(\bar{x} - a)^{n-2}}{\Gamma(n-1)} e^{n(\eta - \bar{x})/\theta} {\bf 1}_{(\bar{x} \geq a \geq \eta)}.$