How do I get the flat tuple type of a tuple?

For this you will need recursive conditional types. This will be fully supported in 4.1:

type Example = [[3,5,7], [4,9], [0,1,10,9]];
type Flatten<T extends any[]> = 
    T extends [infer U, ...infer R] ? U extends any[] ? [...U, ... Flatten<R>]: []: []
type FlatExample = Flatten<Example>;

playground link

edit

A simpler version, courtesy of jcalz:

type Example = [[3,5,7], [4,9], [0,1,10,9]];
type Flatten<T extends any[]> = 
    T extends [any, ...infer R] ? [...T[0], ... Flatten<R>]:  []
type FlatExample = Flatten<Example>;

playground link

/edit

You can even hack a version today (requires extra indirection to trick the compiler into allowing recursive conditional types, which is conceptually equivalent to the simpler version above):

type Example = [[3, 5, 7], [4, 9], [0, 1, 10, 9]];
type Flatten<T extends any[]> = T extends [infer U, ...infer R] ? {
    1: U extends any[] ? [...U, ...Flatten<R>] : [],
    2: []
}[U extends any[] ? 1 : 2] : [];

type FlatExample = Flatten<Example>;

playground link

Just for fun, it's the 4.1 generalized flattened version.

type Example = [[3,5,7], [4,9, [10, 12, [10, 12]]], [0,1,10,9, [10, 12]]];
type Flatten<T extends any[]> = 
    T extends [infer U, ...infer R] ? 
        U extends any[] ? 
        [...Flatten<U>, ... Flatten<R>]: [U, ... Flatten<R>]: []
type FlatExample = Flatten<Example>;

playground link

Note : While recursive types are more supported in 4.1, you can still run into compiler-hardcoded limitations such as type instantiation depth and total type instances (as such recursive types generate many type instantiations). Therefore, use it with caution.