Thanks for providing a definition. I'll go ahead and assume that we're working with vector spaces defined over $\mathbb{C}$, but most of what I say applies equally well over subfields over other archimedean valued fields (in particular, over subfields of $\mathbb{C}$). The short answer is: the usual definition $p(v) = \langle v,v\rangle^{1/2}$ does not produce a seminorm, but modifications such as $p(v) = |\langle v,v\rangle|^{1/2}$ do.
More precisely, let $(V,\langle\cdot,\cdot\rangle)$ denote an indefinite inner product space, in the sense you gave in the comments and for each $v \in V$ define $p(v) = \langle v,v\rangle^{1/2}$. Then it is shown, for instance, here, that $p$ is absolutely homogeneous ($p(\alpha v) = |\alpha|p(v)$ for all $\alpha \in \mathbb{C}$ and $v \in V$) using the usual follow-your-nose proof from the definition of $\langle\cdot,\cdot\rangle$. The "classical" proofs of the other properties required for $p$ to be a seminorm depend on $\langle v,v\rangle \in [0,\infty)$ for all $v \in \mathbb{R}$, which is not the case in general for an indefinite inner product. In fact, the even bigger problem is that this definition of $p$ might not even be well-defined, since $\langle v,v\rangle$ can be negative.
All of these problems do, however, go away if we set $p(v) = |\langle v,v\rangle|^{1/2}$ instead, and the usual proofs of positive semi-definiteness and subadditivity apply, proving that this modified $p$ is a seminorm.
More on indefinite inner products (and links to textbooks covering them) can be found in this older Math.SE thread.
Are you referring to an indefinite inner product as defined on the following Wikipedia page?
https://en.wikipedia.org/wiki/Indefinite_inner_product_space
If so, I can provide an answer later. (In fact, one can glean an answer from the first section of that article!)
– Dan Dec 21 '16 at 20:21