For each $n\ge 1$ there are exactly two extraspecial groups of order $2^{1+2n}$

At $n=1$ an extraspecial group of order eight is nonabelian, hence isomorphic to $\mathrm{Dih}(C_4)$ or to $Q_8$; these have $t=6=2^2+2^1$ and $t=2=2^2-2^1$, so there are exactly two and the value of $t$ tells them apart.

Assume, for every $m$ with $1\le m<n$: both values $2^{2m}+2^m$ and $2^{2m}-2^m$ are realised; every extraspecial group of order $2^{1+2m}$ has an admissible family with at most one quaternion member; if that number is $k$ then $t=2^{2m}+(-1{)}^k{2}^m$; and two such groups with equal $t$ are isomorphic.

If $\varphi :G\to G^{\mathrm{\prime }}$ and $\psi :H\to H^{\mathrm{\prime }}$ are isomorphisms carrying the identified central subgroups to the identified central subgroups compatibly with the identifying isomorphisms, then $\varphi \times \psi$ carries $N$ onto $N^{\mathrm{\prime }}$ and induces an isomorphism $G{\circ }_{\alpha }H\to G^{\mathrm{\prime }}{\circ }_{{\alpha }^{\mathrm{\prime }}}{H}^{\mathrm{\prime }}$; when all four identified subgroups have order two the compatibility is automatic, since a group of order two has only one automorphism.

Let $P$ be extraspecial of order $2^{1+2n}$ with $n\ge 2$ and take an admissible family $P_1,\dots ,P_n$; each member is nonabelian of order eight, hence isomorphic to $\mathrm{Dih}(C_4)$ or to $Q_8$.

If two members $P_i,P_j$ are isomorphic to $Q_8$, then $R=⟨P_i,P_j⟩$ is an internal central product of them, so $R\cong Q_8\circ Q_8$ and $R$ is an internal central product of two subgroups isomorphic to $\mathrm{Dih}(C_4)$ with the same centre $Z(P)$; replacing $P_i,P_j$ by those two subgroups leaves an admissible family with two fewer quaternion members. Repeating, $P$ has an admissible family with $k\in \{0,1\}$ quaternion members.

Fix such a family. Since $n\ge 2$ and $k\le 1$, some member $P_1$ is isomorphic to $\mathrm{Dih}(C_4)$; peeling it leaves $C=⟨P_2,\dots ,P_n⟩$, extraspecial of order $2^{1+2(n-1)}$ with an admissible family of $n-1$ members of which $k$ are quaternion, and $P\cong P_1{\circ }_{\mathrm{id}}C$ along the identity of $Z(P)$.

By the induction hypothesis $t(C)=2^{2(n-1)}+(-1{)}^k{2}^{n-1}$, and $t(P_1)=6=2^2+2$; the counting formula then gives $t(P)=2^{2n}+(-1{)}^k{2}^n$.

If $P$ and $P^{\mathrm{\prime }}$ are extraspecial of order $2^{1+2n}$ with $t(P)=t(P^{\mathrm{\prime }})$, their normalised families have the same $k$ by step 4.1, so peeling a dihedral member from each gives $C$ and $C^{\mathrm{\prime }}$ extraspecial of order $2^{1+2(n-1)}$ with equal $t$, hence isomorphic by the induction hypothesis, by an isomorphism carrying $Z(C)=Z(P)$ onto $Z(C^{\mathrm{\prime }})=Z(P^{\mathrm{\prime }})$; the peeled members are isomorphic too, so $P\cong P^{\mathrm{\prime }}$.

Both values are realised: if $C$ is extraspecial of order $2^{1+2(n-1)}$ then $\mathrm{Dih}(C_4)\circ C$ is extraspecial of order $2^{1+2n}$ with $t=2^{2n}+(-1{)}^k{2}^n$ where $t(C)=2^{2(n-1)}+(-1{)}^k{2}^{n-1}$, so the two groups supplied by the induction hypothesis produce one group of each value. With step 5.1 this gives exactly two isomorphism classes at order $2^{1+2n}$ and completes the induction.