Total number operator and second quantization Let
$$N(\textbf{p}) = \frac{1}{2E_\textbf{p}V}a^{\dagger}(\textbf{p})a(\textbf{p})$$
be the number operator for particles with moment $\textit{p}$ in the volume $\textit{V}$. Then,  one can define
$$N = \sum_{\textbf{p}}a^{\dagger}(\textbf{p})a(\textbf{p})$$
as the total number (of particles) operator. Let
$$|n(\textbf{p}_{1}),n(\textbf{p}_{2}),...\rangle = \frac{\left(a^{\dagger}(\textbf{p}_{1})\right)^{n(\textbf{p}_{1})}\left(a^{\dagger}(\textbf{p}_{2})\right)^{n(\textbf{p}_{2})}...}{\sqrt{\left(2E_{\textbf{p}_{1}}V\right)^{n(\textbf{p}_{1})}n(\textbf{p}_{1})!\left(2E_{\textbf{p}_{2}}V\right)^{n(\textbf{p}_{2})}n(\textbf{p}_{2})!...}} |0 \rangle $$
be a general state. I want to show that, acted upon by the $N$ operator, this state yields a eigenvalue
$$N = n(\textbf{p}_1)+n(\textbf{p}_2)...$$
However, since
$$N(\textbf{p}) |n(\textbf{p})\rangle = n(\textbf{p})|n(\textbf{p})\rangle$$,
my eigenvector is $2E_\textbf{p}V (n(\textbf{p}_1)+n(\textbf{p}_2)...)$ . What am I doing wrong?
Edit
I am using a short-hand notation
$$\sum_{\textbf{p}} = \int \frac{d^3p}{(2\pi)^3 2E_\textbf{p}}$$
 A: The key realization is that
$$a^\dagger(\mathbf{p})a(\mathbf{p})|n(\mathbf{p})\rangle=n(\mathbf{p})|n(\mathbf{p})\rangle.$$ Next, we must assume that $[a(\mathbf{p}_i),a^\dagger(\mathbf{p}_j)]=0$ for $i\neq j$. Then we can generalize our first equation to
$$a^\dagger(\mathbf{p}_i)a(\mathbf{p}_i)|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots,n(\mathbf{p}_i),\cdots\rangle=n(\mathbf{p}_i)|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots,n(\mathbf{p}_i),\cdots\rangle.$$ Your desired relation immediately holds:
$$N|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots\rangle=\sum_i  a^\dagger(\mathbf{p}_i)a(\mathbf{p}_i)|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots\rangle=\sum_i  n(\mathbf{p}_i)|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots\rangle.$$ This is an eigenvalue equation for $N$ with eigenstate $|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots\rangle$ and eigenvalue $\sum_i  n(\mathbf{p}_i)$ [in your notation, the eigenvalue is $\sum_{\mathbf{p}} n(\mathbf{p})$].
EDIT (making everything continuous):
Relativistically-normalized creation operators take the form (up to factors of $2\pi$) $$a^\dagger(\mathbf{p})=(2\pi)^{3/2}\sqrt{2E_{\mathbf{p}}}a^\dagger_{\mathbf{p}}.$$ These have the commutation relations $$[a(\mathbf{p}_1),a^\dagger(\mathbf{p}_2)]=(2\pi)^3 2E_{\mathbf{p}_1}\delta(\mathbf{p_1}-\mathbf{p}_2).$$ Using the commutation relation $n(\mathbf{p}_1)$ times until the annihilation operator acts on the vacuum, our first equation becomes
$$a^\dagger(\mathbf{p})a(\mathbf{p})|n(\mathbf{p}_1)\rangle=(2\pi)^3 2 E_{\mathbf{p}}n(\mathbf{p}_1)\delta(\mathbf{p}-\mathbf{p}_1)|n(\mathbf{p}_1)\rangle,$$ where we explicitly write the delta function and have not integrated over $\mathbf{p}$. We can also be explicit with states that have more than one possible momentum value:
$$a^\dagger(\mathbf{p})a(\mathbf{p}) a^\dagger(\mathbf{p}_1)^{n_1}a^\dagger(\mathbf{p}_2)^{n_2}|{0}\rangle=\left[a^\dagger(\mathbf{p})a^\dagger(\mathbf{p}_1)^{n_1}a(\mathbf{p}) +n_1 a^\dagger(\mathbf{p})a^\dagger(\mathbf{p}_1)^{n_1-1}(2\pi)^3 2E_{\mathbf{p}_1}\delta(\mathbf{p}-\mathbf{p}_1) \right]a^\dagger(\mathbf{p}_2)^{n_2}|{0}\rangle=\left[a^\dagger(\mathbf{p})a^\dagger(\mathbf{p}_1)^{n_1}a^\dagger(\mathbf{p}_2)^{n_2}a(\mathbf{p}) +n_1 a^\dagger(\mathbf{p})a^\dagger(\mathbf{p}_1)^{n_1-1}a^\dagger(\mathbf{p}_2)^{n_2}(2\pi)^3 2E_{\mathbf{p}_1}\delta(\mathbf{p}-\mathbf{p}_1)+n_2 a^\dagger(\mathbf{p})a^\dagger(\mathbf{p}_1)^{n_1}a^\dagger(\mathbf{p}_2)^{n_2-1}(2\pi)^3 2E_{\mathbf{p}_2}\delta(\mathbf{p}-\mathbf{p}_2) \right]|{0}\rangle=\left[n_1 a^\dagger(\mathbf{p})a^\dagger(\mathbf{p}_1)^{n_1-1}a^\dagger(\mathbf{p}_2)^{n_2}(2\pi)^3 2E_{\mathbf{p}_1}\delta(\mathbf{p}-\mathbf{p}_1)+n_2 a^\dagger(\mathbf{p})a^\dagger(\mathbf{p}_1)^{n_1}a^\dagger(\mathbf{p}_2)^{n_2-1}(2\pi)^3 2E_{\mathbf{p}_2}\delta(\mathbf{p}-\mathbf{p}_2) \right]|{0}\rangle=\left[n_1 a^\dagger(\mathbf{p}_1)^{n_1}a^\dagger(\mathbf{p}_2)^{n_2}(2\pi)^3 2E_{\mathbf{p}_1}\delta(\mathbf{p}-\mathbf{p}_1)+n_2 a^\dagger(\mathbf{p}_1)^{n_1}a^\dagger(\mathbf{p}_2)^{n_2}(2\pi)^3 2E_{\mathbf{p}_2}\delta(\mathbf{p}-\mathbf{p}_2) \right]|{0}\rangle=\left[n_1 (2\pi)^3 2E_{\mathbf{p}_1}\delta(\mathbf{p}-\mathbf{p}_1)+n_2 (2\pi)^3 2E_{\mathbf{p}_2}\delta(\mathbf{p}-\mathbf{p}_2) \right]a^\dagger(\mathbf{p}_1)^{n_1}a^\dagger(\mathbf{p}_2)^{n_2}|{0}\rangle=\left[n_1 (2\pi)^3 2E_{\mathbf{p}}\delta(\mathbf{p}-\mathbf{p}_1)+n_2 (2\pi)^3 2E_{\mathbf{p}}\delta(\mathbf{p}-\mathbf{p}_2) \right]a^\dagger(\mathbf{p}_1)^{n_1}a^\dagger(\mathbf{p}_2)^{n_2}|{0}\rangle.$$
Now the picture starts to come together! If we normalize by the energies and integrate over all momentum, the delta functions will give us exactly what we need.
$$\int \frac{d^3 p}{(2\pi)^3 2E_{\mathbf{p}}}a^\dagger(\mathbf{p})a(\mathbf{p})|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots\rangle=\int \frac{d^3 p}{(2\pi)^3 2E_{\mathbf{p}}}\left[n(\mathbf{p}_1) (2\pi)^3 2E_{\mathbf{p}}\delta(\mathbf{p}-\mathbf{p}_1)+n(\mathbf{p}_2) (2\pi)^3 2E_{\mathbf{p}}\delta(\mathbf{p}-\mathbf{p}_2) +\cdots\right]|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots\rangle=\int d^3 p \left[n(\mathbf{p}_1) \delta(\mathbf{p}-\mathbf{p}_1)+n(\mathbf{p}_2) \delta(\mathbf{p}-\mathbf{p}_2) +\cdots\right]|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots\rangle=\left[n(\mathbf{p}_1)+n(\mathbf{p}_2) +\cdots\right]|n(\mathbf{p}_1),n(\mathbf{p}_2),\cdots\rangle$$ as desired. We do not need to refer to the quantization volume $V$ when doing this procedure; we can always introduce it in both the normalization of the integral and of the delta function.
