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 $n>1/c$.
The OP has rectified the confusion raised by the initial formulation of the their question. The changes invalidate the above answer. Here is an updated answer to the current state of the question.
Let $Y:=|X\cdot v|$, where $v$ is a unit eigenvector corresponding to the eigenvalue $s_1$. Then, $0\le Y\le1$ and $EY^2=s_1$. So, for all $a\in(0,1)$ we have the inequality $$1(Y>a)\ge\frac{Y^2-a^2}{1-a^2},$$ with the equality on the event $\{Y\in\{a,1\}\}$, and hence taking expectations gives $$P(|X\cdot v|>a)=P(Y>a)\ge\frac{\max(0,s_1-a^2)}{1-a^2}.$$
This lower bound on $P(|X\cdot v|>a)$ is exact: It is attained if (i) $a^2\le s_1\le1$ and $(X\cdot v)^2$ only takes values $a^2$ and $1$ (with mean $E(X\cdot v)^2=s_1\in[a^2,1]$) or if (ii) $0\le s_1<a^2$ and $(X\cdot v)^2$ only takes value $s_1$.
