The vast majority of disconnected graphs have a single isolated vertex.
Let $A$ be a nonempty proper subset of $\{1,...,n\}$ of size $a$. Let $s(a)$ be the number of graphs with
$e=\lfloor \frac12 {n \choose 2}\rfloor$ edges which have no edges from $A$ to $A^c$.
We want to count the union of all of these. Inclusion-exclusion works, with the dominant terms coming from when $a=1$.
An upper bound is the sum of $s(a)$ over all $A$ of size at most $n/2$, which is at most
$n ~s(1)$ + ${n\choose 2}s(1)$ + $2^ns(3)$.
To get a lower bound, subtract the number of graphs with no edges connecting $A$ to $A^c$ or edges connecting $B$ to $B^c$ for all disjoint $\{A,B\}$. Denote this by $s(\#A,\#B)$. So, subtract
${n\choose2}s(1,1) + 3^ns(1,2)$ from $n~s(1)$.
The rest should be routine estimates on $s(1)$, $s(2)$, $s(3)$, $s(1,1)$, and $s(1,2)$.
$s(a,b) \le s(a+b)$.
$s(a) = ({n\choose 2} -a(n-a))$ choose $e$.
Let the total number of graphs with $e$ edges be $\#G = s(0)$.
$$s(a)/\#G = \prod_{i=0}^{a(n-a)-1} \frac{\lceil{n\choose2}/2\rceil-i}{{n\choose2}-i}.$$
$s(2)/s(1) \le 2^{-n+3}$.
$s(3)/s(1) \le 2^{-2n+8}$.
The dominant term in both the upper bound and the lower bound is $n~s(1)$.
If I calculated correctly, that's asymptotic to $\frac 2 e n 2^{-n} ~\#G$.
Best Answer
The simplest guess one could make is $\frac{1}{50!} { {50 \choose 2} \choose 150}$. That is, we first count the number of labeled such graphs, then assume that most of them have trivial automorphism group so we can approximately divide out by $50!$ when removing the labels. You can estimate how big this is using Stirling's formula.
Edit: Actually, you can also just ask WolframAlpha to compute this estimate. You get $7.028... \times 10^{131}$, which is apparently within 1% of the true answer according to Brendan McKay.