Here is another method.

Since the uniform distribution on the unit disk is rotationally symmetric (invariant under orthogonal transformations), the problem can be reduced to a random walk problem in $\mathbb{R}$.

Let $Z_1,Z_2,\ldots$ be independent, and identically uniformly distributed on the unit disk, with $Z_i=(X_i,Y_i)$. Clearly $X_1,X_2\ldots$ are i.i..d with density $p(x)=\frac{2}{\pi}\sqrt{1-x^2}1_{[-1,1]}(x)$.

For any rotationally symmetric random vector $V=(V_1,V_2)$ with finite expectation of $|V|:=\sqrt{V_1^2+V_2^2}$ it holds that $$\mathbb{E}(|V|)=\frac{\pi}{2}\mathbb{E}(|V_1|)$$ Applying this to $Z^{(n)}:=Z_1+\ldots+Z_n$ we get that

\begin{align*} \mathbb{E}\big(\frac{Z^{(n)}}{n}\big)&=\frac{1}{n}\frac{\pi}{2}\mathbb{E}|X_1+\ldots + X_n|\\ &=\frac{1}{n}\big(\frac{2}{\pi}\big)^{n-1}\,\int_{-1}^1\ldots\int_{-1}^1|x_1+\ldots+x_n|\prod_{i=1}^n\sqrt{1-x_i^2}\,dx_1\,\ldots dx_n\\ \end{align*}

The integrals can be solved explicitly for $n=1$ and $n=2$, but that seems to be as far as it goes. But (using the central limit theorem and uniform integrability) it is easy to see that $$\mathbb{E}\big(\frac{Z^{(n)}}{\sqrt{n}}\big)\longrightarrow \sqrt{\frac{\pi}{8}}$$

Remarks:
(1) a related problem appeared here An interesting triple integral
(2) for similar considerations see Feller II (1971), p. 30 ff

Here is another method.

Since the uniform distribution on the unit disk is rotationally symmetric (invariant under orthogonal transformations), the problem can be reduced to a random walk problem in $\mathbb{R}$.

Let $Z_1,Z_2,\ldots$ be independent, and identically uniformly distributed on the unit disk, with $Z_i=(X_i,Y_i)$. Clearly $X_1,X_2\ldots$ are i.i..d with density $p(x)=\frac{2}{\pi}\sqrt{1-x^2}1_{[-1,1]}(x)$.

For any rotationally symmetric random vector $V=(V_1,V_2)$ with finite expectation of $|V|:=\sqrt{V_1^2+V_2^2}$ it holds that $$\mathbb{E}(|V|)=\frac{\pi}{2}\mathbb{E}(|V_1|)$$ Applying this to $Z^{(n)}:= |Z_1+\ldots+Z_n|$ we get that

\begin{align*} \mathbb{E}\big(\frac{Z^{(n)}}{n}\big)&=\frac{1}{n}\frac{\pi}{2}\mathbb{E}|X_1+\ldots + X_n|\\ &=\frac{1}{n}\big(\frac{2}{\pi}\big)^{n-1}\,\int_{-1}^1\ldots\int_{-1}^1|x_1+\ldots+x_n|\prod_{i=1}^n\sqrt{1-x_i^2}\,dx_1\,\ldots dx_n\\ \end{align*}

The integrals can be solved explicitly for $n=1$ and $n=2$, but that seems to be as far as it goes. But (using the central limit theorem and uniform integrability) it is easy to see that $$\mathbb{E}\big(\frac{Z^{(n)}}{\sqrt{n}}\big)\longrightarrow \sqrt{\frac{\pi}{8}}$$

Remarks:
(1) a related problem appeared here An interesting triple integral
(2) for similar considerations see Feller II (1971), p. 30 ff