Biggest Circle you can fit in a Hypercube

Note that if you take the cross section of a unit cube by a plane perpendicular to the diagonal joining opposite corners you get a regular hexagon of side $\frac {\sqrt 2} 2$. The inscribed circle has diameter $\frac {\sqrt 6}2$.

I would suggest this as a starting point for constructions in higher dimensions.

To make calculations easier, let's take a hypercube of side length $2$ centered at the origin; as is easily verified, the maximal radius of a circle in that cube equals the maximal diameter of a circle in an unit cube.

For even dimension, there's an easy to find circle with radius $\sqrt{n/2}$: $$\{(\underbrace{\cos\phi,\ldots,\cos\phi}_{\frac{n}{2}}, \underbrace{\sin\phi,\ldots,\sin\phi}_{\frac{n}{2}})|0\le\phi\le 2\pi\}$$

This already exceeds your conjecture, as for $n>2$ we have: $$\sqrt{\frac{n}{2}}=\sqrt{\frac{2n}{4}} > \sqrt{\frac{2n}{n+2}}\quad .$$

Interestingly, the "hexagon circle" described by Mark Bennet for the cube ($n=3$) also fits that formula, as then $$\sqrt{\frac{3}{2}} = \sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}\quad.$$

However I suspect that for $n>3$ this is not the biggest circle you can find.

Here's another thought: The longest diameter in an $k$-dimensional hypercube is the diagonal of length $\sqrt{k}$ Now a hypercube with dimension $n=mk$ can be produced as Cartesian product of $m$ hypercubes of dimension $k$. The diagonals then give an inscribed hypercube of dimension $m$ and side length $\sqrt{k}$. Obviously any circle that can be fit in that inscribed hypercube can also be fit into the original hypercube. Therefore if we denote with $d(n)$ the maximal diameter of a circle that can be inscribed in an $n$-dimensional unit hypercube, then we have the relation $$d(mk) \ge\max\{d(m)\,\sqrt{k},\sqrt{m}\,d(k)\}$$

In particular, if $d(k)\ge \sqrt{k/2}$ then $d(mk)\ge \sqrt{mk/2}$.

Edit: In a comment to another answer, Michael Behrend linked to a paper that proves (under the reasonable assumption that the center of the maximal circle must be the center of the cube) that indeed the maximal radius of a circle inscribed in a unit hypercube is $\sqrt{n/8}$, which means the corresponding diameter is $\sqrt{n/2}$, in agreement with the lower bound implied by this answer for any dimension that is a multiple of $2$ or $3$.

This was essentially the Putnam 2008 B3 problem. Here's a link to the solutions for that test: link. It is proven there that for the more general problem of trying put a $k$ dimension ball inside an $n$ dimensional unit cube, the maximum radius of the ball is $\frac{1}{2}\sqrt{\frac{n}{k}}$.