Remove build script?

I've been considering if we could get rid of the build script for determining the host architecture, using [`target_*` cfgs](https://doc.rust-lang.org/reference/conditional-compilation.html) instead of parsing the `TARGET`. Something like the following:

```rust
// Architecture
if cfg!(target_arch = "aarch64") {
Architecture::Aarch64(if cfg!(target_endian = "big") { Aarch64Architecture::Aarch64be } else { Aarch64Architecture::Aarch64 })
} else if cfg!(target_arch = "x86_64") {
Architecture::X86_64
} else if ... // etc.

// Vendor
if cfg!(target_vendor = "unknown") {
Vendor::Amd
} else if cfg!(target_vendor = "amd") {
Vendor::Amd
} else if ... // etc.
```

This would likely allow `target-lexicon` [to be used in `cc`](https://github.com/rust-lang/cc-rs/issues/1219), which in turn would allow us to centralize target parsing even more, to the benefit of all in the ecosystem.

---

One complication though is that Rust's `target_*` cfgs aren't as descriptive, for example `target_arch = "arm"` is often used for several ARM architectures.

I _think_ this may turn out to not really matter though, since [there are fewer targets with host tool support](https://doc.rust-lang.org/nightly/rustc/platform-support.html)? But then again, this would not allow us to distinguish between `armv6-unknown-freebsd` and `armv7-unknown-freebsd`, which, although tier 3, seemingly have host tool support.

There are several ways to solve this, including splitting the build script out into a separate crate that is only loaded when really needed, something like `[target.armv7-unknown-freebsd.dependencies] target-lexion-with-build-script = "..."`.