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EDIT: Thanks a lot to Mike Miller for pointing out in the comments significant simplifications to the proof I wrote

Let $C_n$ be the subgroup of $D_{2n}$ consisting of rotations. This is a normal cyclic subgroup of index 2. Hence, it has cohomology of the form $$H^*(C_n;\mathbb{F}_p)=\mathbb{F}_p[a,b]/a^2$$ where $a$ is in degree 1 and $b$ is in degree 2. So now we can deploy the Lyndon-Hochschild-Serre spectral sequence $$H^*(D_{2n}/C_n;\,H^*(C_n;\mathbb{F}_p))\Rightarrow H^*(D_{2n};\mathbb{F}_p)\,.$$ Since $D_{2n}/C_n$ has order prime to $p$, the spectral sequence degenerates and we arrive at $$H^*(D_{2n};\mathbb{F}_p)\cong H^*(C_n;\mathbb{F}_p)^{D_{2n}/C_n}\,.$$ Note that $D_{2n}/C_n\cong \mathbb{Z}/2$ acts on $C_n$ by sending $z$ to $z^{-1}$. To conclude then it's enough to determine the action of $\mathbb{Z}/2$ on $a$ and $b$.

Let us fix a generator $x\in S$. Then a representative for $a$ is given by the cocycle $\varphi(x^k)=k$, so $\sigma\varphi=-\varphi$, and $\sigma a = -a$.

Moreover, by looking at the Serre spectral sequence for $BC_p→BS^1→BS^1$, we see that $b$ is the image of the generator of $H^2(BS^1;\mathbb{F}_p)$ under the inclusion of $C_n$ in $S^1$, and the action of $\mathbb{Z}/2$ extends to $S^1$. By the Hurewicz theorem we have an isomorphism $$\mathrm{Hom}(\pi_2BS^1;\mathbb{F}_p)\cong H^2(S^1;\mathbb{F}_p)\cong H^2(C_n;\mathbb{F}_p)$$ compatible with the action of $\mathbb{Z}/2$. In particular the action is nontrivial (since the action on $\pi_2BS^1\cong \pi_1S^1\cong\mathbb{Z}$ sends $1$ to $-1$), so $\sigma b= - b$.

Finally $$H^*(D_{2n};\mathbb{F}_p)\cong \left(\mathbb{F}_p[a,b]/a^2\right)^\sigma \cong \mathbb{F}_p[ab,b^2]/(ab)^2$$ In particular it is $\mathbb{F}_p$ in degrees congruent to $0$ and $3$ mod 4, and 0 otherwise.

EDIT: Thanks a lot to Mike Miller for pointing out in the comments significant simplifications to the proof I wrote

First suppose that $p$ does not divide $n$. Then we have $$H^*(D_{2n};\mathbb{F}_p)=\begin{cases}\mathbb{F}_p & \textrm{ if }n=0\\ 0 &\textrm{ otherwise}\end{cases}\,.$$ Otherwise, let us suppose that $p$ divides $n$.

Let $C_n$ be the subgroup of $D_{2n}$ consisting of rotations. This is a normal cyclic subgroup of index 2. Hence, it has cohomology of the form $$H^*(C_n;\mathbb{F}_p)=\mathbb{F}_p[a,b]/a^2$$ where $a$ is in degree 1 and $b$ is in degree 2. So now we can deploy the Lyndon-Hochschild-Serre spectral sequence $$H^*(D_{2n}/C_n;\,H^*(C_n;\mathbb{F}_p))\Rightarrow H^*(D_{2n};\mathbb{F}_p)\,.$$ Since $D_{2n}/C_n$ has order prime to $p$, the spectral sequence degenerates and we arrive at $$H^*(D_{2n};\mathbb{F}_p)\cong H^*(C_n;\mathbb{F}_p)^{D_{2n}/C_n}\,.$$ Note that $D_{2n}/C_n\cong \mathbb{Z}/2$ acts on $C_n$ by sending $z$ to $z^{-1}$. To conclude then it's enough to determine the action of $\mathbb{Z}/2$ on $a$ and $b$.

Let us fix a generator $x\in S$. Then a representative for $a$ is given by the cocycle $\varphi(x^k)=k$, so $\sigma\varphi=-\varphi$, and $\sigma a = -a$.

Moreover, by looking at the Serre spectral sequence for $BC_n→BS^1→BS^1$, we see that $b$ is the image of the generator of $H^2(BS^1;\mathbb{F}_p)$ under the inclusion of $C_n$ in $S^1$, and the action of $\mathbb{Z}/2$ extends to $S^1$. By the Hurewicz theorem we have an isomorphism $$\mathrm{Hom}(\pi_2BS^1;\mathbb{F}_p)\cong H^2(BS^1;\mathbb{F}_p)\cong H^2(BC_n;\mathbb{F}_p)$$ compatible with the action of $\mathbb{Z}/2$. In particular the action is nontrivial (since the action on $\pi_2BS^1\cong \pi_1S^1\cong\mathbb{Z}$ sends $1$ to $-1$), so $\sigma b= - b$.

Finally $$H^*(D_{2n};\mathbb{F}_p)\cong \left(\mathbb{F}_p[a,b]/a^2\right)^\sigma \cong \mathbb{F}_p[ab,b^2]/(ab)^2$$ In particular it is $\mathbb{F}_p$ in degrees congruent to $0$ and $3$ mod 4, and 0 otherwise.

Let $S$ be the $p$-Sylow subgroup of $D_{2n}$ for $p$ odd. Then $S$ is normal and cyclic. Moreover it has cohomology of the form $$H^*(S;\mathbb{F}_p)=\mathbb{F}_p[a,b]/a^2$$ where $a$ is in degree 1 and $b$ is in degree 2. So now we can deploy the Lyndon-Hochschild-Serre spectral sequence $$H^*(D_{2n}/S;\,H^*(S;\mathbb{F}_p))\Rightarrow H^*(D_{2n};\mathbb{F}_p)\,.$$ Since $D_{2n}/S$ has order prime mod $p$, the spectral sequence degenerates and we arrive at $$H^*(D_{2n};\mathbb{F}_p)\cong H^*(S;\mathbb{F}_p)^{D_{2n}/S}\cong\left(\mathbb{F}_p[a,b]/a^2\right)^{D_{2n}}\,.$$ To conclude then it's enough to determine the action of $D_{2n}$ on $a$ and $b$. Since the centralizer of $S$ is the subgroup of rotations $C_n<D_{2n}$, we only have to worry about reflections. Let $\sigma$ be a reflection, so that its action on $S$ sends $x$ to $x^{-1}$.

Let us fix a generator $x\in S$. Then a representative for $a$ is given by the cocycle $\varphi(x^k)=k$, so $\sigma\varphi=-\varphi$, and $\sigma a = -a$.

Moreover, by looking at the Serre spectral sequence for $BC_p→BS^1→BS^1$, we see that $b$ is the image of the generator of $H^2(BS^1;\mathbb{F}_p)$ under the inclusion of $S$ in $S^1$, and so in particular the map $H^2(S;\mathbb{F}_p)\to H^2(C_p;\mathbb{F}_p)$ is an isomorphism. So it is enough to understand the action when $S=C_p$. But in this case $b$ is the Bockstein of $a$, and so $\sigma b= - b$. Finally $$H^*(D_{2n};\mathbb{F}_p)\cong \left(\mathbb{F}_p[a,b]/a^2\right)^\sigma \cong \mathbb{F}_p[ab,b^2]/(ab)^2$$ In particular it is $\mathbb{F}_p$ in degrees congruent to $0$ and $3$ mod 4, and 0 otherwise.

EDIT: Thanks a lot to Mike Miller for pointing out in the comments significant simplifications to the proof I wrote

Let $C_n$ be the subgroup of $D_{2n}$ consisting of rotations. This is a normal cyclic subgroup of index 2. Hence, it has cohomology of the form $$H^*(C_n;\mathbb{F}_p)=\mathbb{F}_p[a,b]/a^2$$ where $a$ is in degree 1 and $b$ is in degree 2. So now we can deploy the Lyndon-Hochschild-Serre spectral sequence $$H^*(D_{2n}/C_n;\,H^*(C_n;\mathbb{F}_p))\Rightarrow H^*(D_{2n};\mathbb{F}_p)\,.$$ Since $D_{2n}/C_n$ has order prime to $p$, the spectral sequence degenerates and we arrive at $$H^*(D_{2n};\mathbb{F}_p)\cong H^*(C_n;\mathbb{F}_p)^{D_{2n}/C_n}\,.$$ Note that $D_{2n}/C_n\cong \mathbb{Z}/2$ acts on $C_n$ by sending $z$ to $z^{-1}$. To conclude then it's enough to determine the action of $\mathbb{Z}/2$ on $a$ and $b$.

Let us fix a generator $x\in S$. Then a representative for $a$ is given by the cocycle $\varphi(x^k)=k$, so $\sigma\varphi=-\varphi$, and $\sigma a = -a$.

Moreover, by looking at the Serre spectral sequence for $BC_p→BS^1→BS^1$, we see that $b$ is the image of the generator of $H^2(BS^1;\mathbb{F}_p)$ under the inclusion of $C_n$ in $S^1$, and the action of $\mathbb{Z}/2$ extends to $S^1$. By the Hurewicz theorem we have an isomorphism $$\mathrm{Hom}(\pi_2BS^1;\mathbb{F}_p)\cong H^2(S^1;\mathbb{F}_p)\cong H^2(C_n;\mathbb{F}_p)$$ compatible with the action of $\mathbb{Z}/2$. In particular the action is nontrivial (since the action on $\pi_2BS^1\cong \pi_1S^1\cong\mathbb{Z}$ sends $1$ to $-1$), so $\sigma b= - b$. 

Finally $$H^*(D_{2n};\mathbb{F}_p)\cong \left(\mathbb{F}_p[a,b]/a^2\right)^\sigma \cong \mathbb{F}_p[ab,b^2]/(ab)^2$$ In particular it is $\mathbb{F}_p$ in degrees congruent to $0$ and $3$ mod 4, and 0 otherwise.

Let $S$ be the $p$-Sylow subgroup of $D_{2n}$ for $p$ odd. Then $S$ is normal and cyclic. Moreover it has cohomology of the form $$H^*(S;\mathbb{F}_p)=\mathbb{F}_p[a,b]/a^2$$ where $a$ is in degree 1 and $b$ is in degree 2. So now we can deploy the Lyndon-Hochschild-Serre spectral sequence $$H^*(D_{2n}/S;\,H^*(S;\mathbb{F}_p))\Rightarrow H^*(D_{2n};\mathbb{F}_p)\,.$$ Since $D_{2n}/S$ has order prime mod $p$, the spectral sequence degenerates and we arrive at $$H^*(D_{2n};\mathbb{F}_p)\cong H^*(S;\mathbb{F}_p)^{D_{2n}/S}\cong\left(\mathbb{F}_p[a,b]/a^2\right)^{D_{2n}}\,.$$ To conclude then it's enough to determine the action of $D_{2n}$ on $a$ and $b$. Since the centralizer of $S$ is the subgroup of rotations $C_n<D_{2n}$, we only have to worry about reflections. Let $\sigma$ be a reflection, so that its action on $S$ sends $x$ to $x^{-1}$.

Let us fix a generator $x\in S$. Then a representative for $a$ is given by the cocycle $\varphi(x^k)=k$, so $\sigma\varphi=-\varphi$, and $\sigma a = -a$.

Moreover, by looking at the Serre spectral sequence for $BC_p→BS¹→BS¹$, we see that $b$ is the image of the generator of $H^2(S_1;\mathbb{F}_p)$ under the inclusion of $S$ in $S^1$, and so in particular the map $H^2(S;\mathbb{F}_p)\to H^2(C_p;\mathbb{F}_p)$ is an isomorphism. So it is enough to understand the action when $S=C_p$. But in this case $b$ is the Bockstein of $a$, and so $\sigma b= - b$. Finally $$H^*(D_{2n};\mathbb{F}_p)\cong \left(\mathbb{F}_p[a,b]/a^2\right)^\sigma \cong \mathbb{F}_p[ab,b^2]/(ab)^2$$ In particular it is $\mathbb{F}_p$ in degrees congruent to $0$ and $3$ mod 4, and 0 otherwise.

Let $S$ be the $p$-Sylow subgroup of $D_{2n}$ for $p$ odd. Then $S$ is normal and cyclic. Moreover it has cohomology of the form $$H^*(S;\mathbb{F}_p)=\mathbb{F}_p[a,b]/a^2$$ where $a$ is in degree 1 and $b$ is in degree 2. So now we can deploy the Lyndon-Hochschild-Serre spectral sequence $$H^*(D_{2n}/S;\,H^*(S;\mathbb{F}_p))\Rightarrow H^*(D_{2n};\mathbb{F}_p)\,.$$ Since $D_{2n}/S$ has order prime mod $p$, the spectral sequence degenerates and we arrive at $$H^*(D_{2n};\mathbb{F}_p)\cong H^*(S;\mathbb{F}_p)^{D_{2n}/S}\cong\left(\mathbb{F}_p[a,b]/a^2\right)^{D_{2n}}\,.$$ To conclude then it's enough to determine the action of $D_{2n}$ on $a$ and $b$. Since the centralizer of $S$ is the subgroup of rotations $C_n<D_{2n}$, we only have to worry about reflections. Let $\sigma$ be a reflection, so that its action on $S$ sends $x$ to $x^{-1}$.

Let us fix a generator $x\in S$. Then a representative for $a$ is given by the cocycle $\varphi(x^k)=k$, so $\sigma\varphi=-\varphi$, and $\sigma a = -a$.

Moreover, by looking at the Serre spectral sequence for $BC_p→BS^1→BS^1$, we see that $b$ is the image of the generator of $H^2(BS^1;\mathbb{F}_p)$ under the inclusion of $S$ in $S^1$, and so in particular the map $H^2(S;\mathbb{F}_p)\to H^2(C_p;\mathbb{F}_p)$ is an isomorphism. So it is enough to understand the action when $S=C_p$. But in this case $b$ is the Bockstein of $a$, and so $\sigma b= - b$. Finally $$H^*(D_{2n};\mathbb{F}_p)\cong \left(\mathbb{F}_p[a,b]/a^2\right)^\sigma \cong \mathbb{F}_p[ab,b^2]/(ab)^2$$ In particular it is $\mathbb{F}_p$ in degrees congruent to $0$ and $3$ mod 4, and 0 otherwise.