Effective security of block cipher – equal the key size, or half the key size?
I just read that chapter of the book, and the authors don't really justify their claim.

They also talk about "using random data to prevent collision and precomputation attacks" (which would then give you back the full key-size crypto strength) – this is about using random initialization vectors and such.

But if you are using an insecure mode of operation, or a non-random IV where a random one is required, using a bigger key-size doesn't really help, as the attacks possible here are not really key-recovery attacks.

Where such stuff could work would be for hash functions (which have at most $2^{n/2}$ tries collision resistance, assuming $n$-bit output), and a secret key included in the hash makes the brute-forcing impossible. But this certainly is not "symmetric encryption".

Another possibility which would reduce complexity to half the key-size would be assuming a (large enough) quantum computer – that could do a brute-force key search in $2^{n/2}$ steps instead of $2^{n}$ steps. But the text in the chapter doesn't really look like they are referring to this.

The crypto strength of an ideal symmetric cipher in absence of new generic attacks is equal to its key size (i.e. you need in average $2^n/2$ tries for a brute force key search attack).

Generally this is valid for all non-broken encryption algorithms, too (just by definition of "non-broken"): A symmetric cipher with key-size $n$ bit should offer $n$ bit crypto strength.

So my say would be: "Use a bigger key-size if you are concerned about future quantum computer attacks (or generally attacks on the algorithm which shave off some of the security, but not all of it), otherwise stay at 128 bit."