The answer is no: in general (and usually) there are no positive absolute constants $a$ and $b$ such that for some unit vector $v$ one has $$P(|X\cdot v|\ge a)\ge b.$$
Indeed, otherwise one would have $E(X\cdot v)^2\ge c:=ba^2>0$. However, if $X$ is uniformly distributed on the unit sphere in $\mathbb R^n$ and $v$ is a unit vector, then $(X\cdot v)^2$ has the beta distribution with parameters $1/2,(n-1)/2$ and hence $E(X\cdot v)^2=1/n<c$ if $n>1/c$.