I doubt that you'll find an exact closed form. However, since the event duration $Y$ is measured in milliseconds and the time period $T$ in hours, it seems likely that for practical purposes only the leading term in an expansion in $q:=Y/T$ will be relevant.
In that case, the possibility of more than $Z$-fold overlap can be neglected, and the desired probability is just $\binom XZ$ times the probability $p$ that $Z$ uniformly distributed events overlap. Boundary effects will also be of higher order in $Y/T$ than the leading term, so we can neglect them and treat the time period as if it wrapped around at the ends.
Here are two different ways to determine $p$, and also the probability distribution of the lengths of the overlaps.
Let $p(L)$ be the probability density of an overlap of length $L$, that is, $p(L)\mathrm dL$ is the probability of an overlap of length between $L$ and $L+\mathrm dL$. For any given instant $t$ in $T$, the probability that the interval $[t,t+L]$ is entirely covered by a given one of the events is $(Y-L)/T$, so the probability that it is covered by all $Z$ events is $((Y-L)/T)^Z$. Integrating this over $t$ gives the measure of points followed by a fully covered interval of length $L$. But we can also obtain the same measure in another way: An overlap of length $L'\gt L$ leads to an interval of length $L'-L$ of points followed by a fully covered interval of length $L$. Integrating this over $L'$ from $L$ to $Y$ also gives the measure of points followed by a fully covered interval of length $L$. Thus we have
$\int_L^Yp(L')(L'-L)\,\mathrm dL'=\int_0^T\left(\frac{Y-L}T\right)^Z\,\mathrm dt=T\left(\frac{Y-L}T\right)^Z\;.$
Applying $-\mathrm d/\mathrm dL$ yields
$\int_L^Yp(L')\,\mathrm dL'=Z\left(\frac{Y-L}T\right)^{Z-1}\;,\tag{1}$
and then setting $L=0$ gives
$p=Z\left(\frac YT\right)^{Z-1}\;.$
To get the distribution of the overlap lengths, we can apply $-\mathrm d/\mathrm dL$ once more to $(1)$ to obtain
$p(L)=\frac{Z(Z-1)}T\left(\frac{Y-L}T\right)^{Z-2}\;.$
Another way to derive this result is to place the events one by one and keep track of the probability that an overlap of length $L$ with $0\lt L\le Y$ remains. With one event placed, the "overlap" is $L$ with probability $1$. If we place a second event, there's a window of size $2Y$ in which an overlap will occur, so an overlap will occur with probability $2q$, and its length ranges from $0$ to $Y$ with uniform probability, so the probability density is $2q/Y=2/T$ for $0\lt L\le Y$.
Generally, if there was an overlap of length $L'$ and we place another event, there's a probability $(Y-L')/T$ that the same overlap remains, and a probability $2L'/T$ that the overlap is reduced to a new value $L\gt0$, with uniform probability for all values $L\lt L'$. This gives a recurrence for the probability density $p_Z(L)$:
$p_Z(L)=\frac{Y-L}Tp_{Z-1}(L)+\frac2T\int_L^Yp_{Z-1}(L')\,\mathrm dL'\;.$
It's straighforward to check that the result derived above satisfies this recurrence with the initial condition $p_2(L)=2/T$.