For the continuous version, assume that you pick $n$ points randomly from $[0, 1]^2$. You are trying to find $$E\left[ \frac{1}{n}\sum_{k=1}^n \min_{i, 1 \le i \le n, i \not=k}||X_k-X_i|| \right] \tag 1$$
where each of $X_i$ are randomly and uniformly chosen from $[0, 1]^2$.
Because the expected value of a sum of random variables is equal to the sum of the expected values, this simplifies to $$ \frac{1}{n}\sum_{k=1}^n E\left[\min_{i, 1 \le i \le n, i \not=k}||X_k-X_i|| \right] \tag 2$$
By symmetry, this is $$E\left[\min_{i, 1 \le i \le n-1}||X_n-X_i|| \right] \tag 3$$ Let $F_n(x) = P\left(\min_{i, 1 \le i \le n-1}||X_n-X_i|| > x\right)$, which would be $1$ for $x < 0$ and $0$ for $x > \sqrt{2}$. Then $(3)$ is $\int_0^{\sqrt{2}}F_n(x) dx$. $F_n(x)$ is equal to $$P\left(||X_n-X_1|| > x\right)^{n-1} \tag 4$$
The probability $P\left(||X_n-X_1|| > x\right)$ is given using formula $(6)$ from this paper as $\int_x^{\sqrt{2}}g(t)dt$ where $$g(x) = \begin{cases}
2x^{3}-8x^{2}+2\pi x & 0 \leq x < 1 \\
-2x^{3}-\left(2\pi+4\right)x+8x\operatorname{arccsc}\left(x\right)+8x\sqrt{x^{2}-1} & 1\leq x < \sqrt{2}
\end{cases} \tag 5$$
Therefore, $P\left(||X_n-X_1|| > x\right)$ is given by $$\begin{cases}
-\frac{x^{4}}{2}+\frac{8x^{3}}{3}-\pi x^{2}+1 & 0 \leq x < 1 \\
\frac{2}{3}-\frac{4}{3}\left(2x^{2}+1\right)\sqrt{x^{2}-1}+\frac{1}{2}x^{4}+\left(2+\pi\right)x^{2}-4x^{2}\operatorname{arccsc}\left(x\right) & 1 \leq x < \sqrt{2}
\end{cases} \tag 6$$
The integral $\int_0^{\sqrt{2}}P\left(||X_n-X_1|| > x\right)^{n-1}dx$ is unlikely to have a closed form (or if it does, it would probably be very complicated). Instead, for an approximation, split it up as $$\int_0^{1}P\left(||X_n-X_1|| > x\right)^{n-1}dx + \int_1^{\sqrt{2}}P\left(||X_n-X_1|| > x\right)^{n-1}dx$$
The second integral has an upper bound of $\left(\frac{19}{6}-\pi\right)^{n-1}\left(\sqrt{2}-1\right)$, obtained because each value in that interval has an upper bound of $P\left(||X_n-X_1|| > 1\right)^{n-1} = \left(\frac{19}{6}-\pi\right)^{n-1}$. The first integral is given by $$\sum_{k=0}^{n-1}\sum_{m=0}^{n-k-1}\sum_{r=0}^{n-k-1}\frac{\binom{n-1}{k, m, r}}{\left(1+2k+3m+4r\right)}\left(-\pi\right)^{k}\left(\frac{8}{3}\right)^{m}\left(-\frac{1}{2}\right)^{r}$$
Numerically, this approaches $\frac{1}{2\sqrt{n-1}}$, but I'm having trouble proving it.