Here are some lower bounds including a wild speculation that the correct value for largish $n$ might be roughly $0.5^n.$ 

The chance that a single line avoids the tilted square with corners $(1/2,0),(1,1/2),(1/2,1),(0,1/2)$ (of area $1/2$) is $1/6:$ With probability $1$, the point first chosen is not a corner of the original square or the tilted one. wlog it is on the left half of the bottom. Then the second point needs to be on the lower half of the right side. The chance of this is 1/6. So the chance that none of the $n$ lines cut the tilted square is $1/(6^n).$

This can be improved to $4/(6^n)=2/(3\cdot 6^{n-1})$ as follows. The chance that the first line crosses two adjacent sides is $2/3.$ If this does happen then one of the diagonals misses this line and splits the square into two isosceles right triangles with area $1/2.$ One is cut and the other intact. For every subsequent choice the first point is outside the intact half with probability $1/2$ and the second with probability $1/3.$

If I did a randomized experiment correctly then a lower bound of about $0.34^n$ results from looking at the chance that that all $n$ lines miss the circle of radius $1/\sqrt{2\pi}$ centered at $(1/2,1/2).$ 

**LATER** Here is a much improved lower bound.
[![enter image description here][1]][1]

If I calculated correctly, the octagon above is the best of its kind. I get that the probability it stays intact is about $(0.4797667)^n.$ The lower left corner is at about $(0.32606,0.11670).$

I think a slight improvement would result from making the horizontal and vertical sides slightly shorter and replacing the tilted sides by something like a circular arc. It would be pleasing to get the lower bound over $0.5^n$ but I don't know if that is true.


  [1]: https://i.sstatic.net/Xn0g8.png