$\newcommand{\RR}{\mathbb{R}}\newcommand{\Trace}{\operatorname{tr}}$ No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and associates, follows.
[sp]The largest possible radius is $\frac{\sqrt{2}}{2}$. It will be convenient to solve the problem for a hypercube of side length 2 instead, in which case we are trying to show that the largest radius is $\sqrt{2}$.
Choose coordinates so that the interior of the hypercube is the set $H = [-1,1]^4$ in $\RR^4$. Let $C$ be a circle centered at the point $P$. Then $C$ is contained both in $H$ and its reflection across $P$; these intersect in a rectangular paralellepiped each of whose pairs of opposite faces are at most 2 unit apart. Consequently, if we translate $C$ so that its center moves to the point $O = (0,0,0,0)$ at the center of $H$,
then it remains entirely inside $H$.
This means that the answer we seek equals the largest possible radius of a circle $C$ contained in $H$ \emph{and centered at $O$}. Let $v_1 = (v_{11}, \dots, v_{14})$ and $v_2 = (v_{21},\dots,v_{24})$ be two points on $C$ lying on radii perpendicular to each other. Then the points of the circle can be expressed as $v_1 \cos(\theta) + v_2 \sin(\theta)$ for $0 \leq \theta < 2\pi$. Then $C$ lies in $H$ if and only if for each $i$, we have
\[
|v_{1i} \cos(\theta) + v_{2i} \sin(\theta)|
\leq 1 \qquad (0 \leq \theta < 2\pi).
\]
In geometric terms, the vector $(v_{1i}, v_{2i})$ in $\RR^2$ has dot product at most 1 with every unit vector. Since this holds for the unit vector in the same direction as $(v_{1i}, v_{2i})$, we must have
\[
v_{1i}^2 + v_{2i}^2 \leq 1 \qquad (i=1,\dots,4).
\]
Conversely, if this holds, then the Cauchy-Schwarz inequality and the above analysis imply that $C$ lies in $H$.
If $r$ is the radius of $C$, then
\begin{align*}
2 r^2 &= \sum_{i=1}^4 v_{1i}^2 + \sum_{i=1}^4 v_{2i}^2 \\
&= \sum_{i=1}^4 (v_{1i}^2 + v_{2i}^2) \\
&\leq 4,
\end{align*}
so $r \leq \sqrt{2}$. Since this is achieved by the circle through $(1,1,0,0)$ and $(0,0,1,1)$, it is the desired maximum.
Remark:
One may similarly ask for the radius of the largest $k$-dimensional ball inside an $n$-dimensional unit hypercube; the given problem is the case $(n,k) = (4,2)$. Daniel Kane gives the following argument to show that the maximum radius in this case is $\frac{1}{2} \sqrt{\frac{n}{k}}$. (Thanks for Noam Elkies for passing this along.)
We again scale up by a factor of 2, so that we are trying to show that the maximum radius $r$ of a $k$-dimensional ball contained in the hypercube $[-1,1]^n$ is $\sqrt{\frac{n}{k}}$. Again, there is no loss of generality in centering the ball at the origin. Let $T: \RR^k \to \RR^n$ be a similitude carrying the unit ball to this embedded $k$-ball. Then there exists a vector $v_i \in \RR^k$ such that for $e_1,\dots,e_n$ the standard basis of $\RR^n$, $x \cdot v_i = T(x) \cdot e_i$ for all $x \in \RR^k$. The condition of the problem is equivalent to requiring $|v_i| \leq 1$ for all $i$, while the radius $r$ of the embedded ball is determined by the fact that for all $x \in \RR^k$,
\[
r^2 (x \cdot x) = T(x) \cdot T(x) = \sum_{i=1}^n x \cdot v_i.
\]
Let $M$ be the matrix with columns $v_1,\dots,v_k$; then $MM^T = r^2 I_k$, for $I_k$ the $k \times k$ identity matrix. We then have
\begin{align*}
kr^2 &= \Trace(r^2 I_k) = \Trace(MM^T)\\
&= \Trace(M^TM) = \sum_{i=1}^n |v_i|^2 \\
&\leq n,
\end{align*}
yielding the upper bound $r \leq \sqrt{\frac{n}{k}}$.
To show that this bound is optimal, it is enough to show that one can find an orthogonal projection of $\RR^n$ onto $\RR^k$ so that the projections of the $e_i$ all have the same norm (one can then rescale to get the desired configuration of $v_1,\dots,v_n$). We construct such a configuration by a ``smoothing'' argument. Start with any projection. Let $w_1,\dots,w_n$ be the projections of $e_1,\dots,e_n$. If the desired condition is not achieved, we can choose $i,j$ such that
\[
|w_i|^2 < \frac{1}{n} (|w_1|^2 + \cdots + |w_n|^2) < |w_j|^2.
\]
By precomposing with a suitable rotation that fixes $e_h$ for $h \neq i,j$, we can vary $|w_i|, |w_j|$ without varying $|w_i|^2 + |w_j|^2$ or $|w_h|$ for $h \neq i,j$. We can thus choose such a rotation to force one of $|w_i|^2, |w_j|^2$ to become equal to $\frac{1}{n} (|w_1|^2 + \cdots + |w_n|^2)$. Repeating at most $n-1$ times gives the desired configuration.
[/sp]
