$\newcommand\R{\mathbb R}\newcommand{\Si}{\Sigma}\newcommand{\si}{\sigma}$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
\begin{equation}
P(|X\cdot v|>a)=P(Y>a)\ge\frac{\max(0,s_1-a^2)}{1-a^2}. \tag{1}\label{1}
\end{equation}
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 $s_1$.
Addendum 1: Strictly speaking, to show that the lower bound on $P(|X\cdot v|>a)$ in \eqref{1} is exact, we also need to show that
(I) for any $s_1\in[a^2,1]$ and at least for some $n$, there exist a random unit vector $X$ in $\R^n$ and a unit vector $v\in\R^n$ such that $(X\cdot v)^2$ only takes values $a^2$ and $1$, with mean $E(X\cdot v)^2=s_1$, and, moreover, $s_1$ is the largest eigenvalue of the covariance matrix $\Si$ of $X$;
(II) for any $s_1\in(0,a^2)$ and at least for some $n$, there exist a random unit vector $X$ in $\R^n$ and a unit vector $v\in\R^n$ such that $(X\cdot v)^2$ only takes value $s_1$ and, moreover, $s_1$ is the largest eigenvalue of the covariance matrix $\Si$ of $X$.
To prove (I), do take any $s_1\in[a^2,1]$, and take any unit vector $v\in\R^n$, where $n\ge2$.
Then let $\mu_X=p\mu_V+\frac q2\,\mu_W+\frac q2\,\mu_{-W}$, where $\mu_Y$ denotes the distribution of a random vector $Y$,
\begin{equation}
p:=\frac{s_1-a^2}{1-a^2},\quad q:=1-p=\frac{1-s_1}{1-a^2},
\end{equation}
$P(V=v)=P(V=-v)=1/2$, $W:=av+\sqrt{1-a^2}\,U$, and $U$ is uniformly distributed on the unit sphere of the vector space that is the orthogonal complement of $\text{span}(\{v\})$ to $\R^n$. Then $P((X\cdot v)^2=1)=p$, $P((X\cdot v)^2=a^2)=q$, $E(X\cdot v)^2=s_1$, and the eigenvalues of the covariance matrix $\Si$ of $X$ are $s_1$ and $\dfrac{1-s_1}{n-1}$ (the latter one of multiplicity $n-1$). So, if $n\ge1/s_1$, then $s_1$ is the largest eigenvalue of $\Si$. Thus, all the desired conditions are satisfied, and (I) is proved.
The proof of (II) is similar, and a bit simpler. Here, we let $\mu_X=\frac12\,\mu_T+\frac12\,\mu_{-T}$, where $T:=\sqrt{s_1}\,v+\sqrt{1-s_1}\,U$.
Addendum 2: The lower bound in your post depends on the variance $\si^2$ of $Z=Y^2$. You did not fully specify a value of $\si^2$, just noting that $\si^2\le1/4$, since $0\le Z\le1$. The latter bound on $\si^2$ can be improved to the optimal (in terms of $EZ$) bound $EZ(1-EZ)=s_1(1-s_1)$, which is attained when $Z$ only takes values $0$ and $1$. It now follows that your lower bound cannot be exact -- because, as seen from above, the exact lower bound is only attained when $Z=(X\cdot v)^2$ takes values in the set $\{a^2,s_1,1\}$, which does not contain the value $0$. Also, it can probably be shown directly that, in distinction from the lower bound on $P(|X\cdot v|>a)$ in \eqref{1}, the lower bound in your post cannot be exact.
