Assume we have a centrally symmetric convex set $K \subset \mathbb{R}^n$ such that Vol(K)=1. In addition, assume that for every direction $u$ we know that $Vol(K \Delta R_u(K)) < \epsilon$, where $A \Delta B$ is the symmetric difference and $R_u(K)$ denotes the reflection of $K$ with respect to $u^\perp$. Does this imply that $K$ is close (in terms of $\epsilon$) to a Euclidean ball (in the same metric)?

Let $r$ be the radius of the maximal ball contained in $K$ and $R$ the radius of the minimal ball containing $K$. I claim that $\frac rR>12\epsilon^{1/n}$. It follows that the Hausdorff distance from $K$ to a ball is bounded by $C_n\epsilon^{1/n}$. The argument does not use symmetry (although the constant can be improved in the symmetric case). Let $v$ and $v'$ be unit vectors such that $Rv$ and $rv'$ belong to the boundary of $K$. Apply a reflection which sends $v'$ to $v$ and let $K'$ be the resulting body. Let $h$ be the homothety centered at $Rv$ with ratio $\frac{Rr}{2R}$. Then $h(K)$ does not intersect the interior of $K'$ because they are separated by the hyperplane through $rv$ orthogonal to $v$. On the other hand, $h(K)\subset K$ due to convexity. Thus $h(K)\subset K\setminus int(K')$, hence $Vol(h(K))\le Vol(K\Delta K')<\epsilon$. But $Vol(h(K))=\left(\frac{Rr}{2R}\right)^n=\frac1{2^n}\left(1\frac rR\right)^n$. Hence $1\frac rR<2\epsilon^{1/n}$ as claimed. Added later. Now let us show that the Nikodym distance from $K$ to a ball is bounded by $O(\epsilon)$. Let $f:S^{n1}\to\mathbb R_+$ be the radial function of $K$. Then $$ Vol(K\Delta R_u(K)) =\frac1n \int_{S^{n1}} f(x)^nf(R_ux)^n dx \sim c(n) \int_{S^{n1}} f(x)f(R_ux) dx . $$ The last equivalence holds because we already know that $f$ is uniformly close to a known constant. Thus $f$ lies within $L^1$ distance $O(\epsilon)$ from every its reflection and hence from any rotation. Averaging over all rotations yields that $$ \int_{S^{n1}}\int_{S^{n1}} f(x)f(y) dxdy \le C\epsilon $$ (where $C$ depends on $n$). This easily implies that $f$ lies within $L^1$ distance $O(\epsilon)$ from a constant. Indeed there is $r_0$ such that the volumes of both sets $A=\{x:f(x)\le r_0\}$ and $B=\{x:f(x)\ge r_0\}$ are at least half the total volume. Restricting the above integral to $x\in A$ and $y\in B$ shows that the integral mean of $f(x)f(y)$ is at least 1/4 of the integral mean of $f(x)r_0$ over $x\in S^{n1}$. Thus $$ \int_{S^{n1}} f(x)r_0 dx \le C_1\epsilon $$ for some $C_1=C_1(n)$. This means that $K$ lies within Nikodym distance $C_2\epsilon$ from the ball of radius $r_0$ and hence within Nikodym distance $C_3\epsilon$ from the ball of volume 1. 


For convex closed sets with uniform upperdiameter and lowervolume bounds, the Nikodym distance is equivalent to Hausdorff distance. Your condition for the Hausdorff distance, implies that $K$ is close to any ball which does not contain $K$ and is not contained in $K$. Hence the statement follows. 


Isn't the diameter bounded by square root of n? 

