The problem is equivalent to determine whether the Fourier coefficient of
$$
\Theta(q)=\sum_{
\begin{array}{cc}
n_1,n_2,n_3,n_4\in\textbf{Z}\\
n_1+n_2+n_3+n_4=1
\end{array}
}q^{n_1^2+n_2^2+n_3^2+n_4^2}
$$
is always greater or equal to 1.
We can write equivalent $n_4=1-n_1-n_2-n_3$.
Hence
$$
\sum^{4}_{j=1}n_j^2=
$$
$$
=1+2n_1+n_2^2+2n_2+2n_1n_2+2n_2^2+2n_3+2n_1n_3+2n_2n_3+2n_3^2=
$$
$$
=(n_1+n_2+n_3)^2+2(n_1+n_2+n_3)+n_1^2+n_2^2+n_3^2+1
$$
Hence
$$
\Theta(q)=\sum_{
\begin{array}{cc}
n_1,n_2,n_3,t\in\textbf{Z}\\
n_1+n_2+n_3=t
\end{array}
}q^{(t+1)^2+n_1^2+n_2^2+n_3^2}=
$$
$$
=\sum^{\infty}_{t=-\infty}q^{(t+1)^2}\sum_{
\begin{array}{cc}
n_1,n_2,n_3\in\textbf{Z}\\
n_1+n_2+n_3=t
\end{array}
}q^{n_1^2+n_2^2+n_3^2}=
$$
$$
=\sum^{\infty}_{t=-\infty}q^{(t+1)^2}\sum^{\infty}_{s=0}q^s\sum_{
\begin{array}{cc}
n_1,n_2,n_3\in\textbf{Z}\\
n_1+n_2+n_3=t\\
n_1^2+n_2^2+n_3^2=s\\
\end{array}
}1.
$$
But the equations
$$
n_1+n_2+n_3=t\textrm{, }n_1^2+n_2^2+n_3^2=s
$$
have solutions
$$
n_1=\frac{1}{2}\left(-n_3+t-\sqrt{-2n_3^2+2s+2n_3t-t^2}\right)
$$
$$
n_2=\frac{1}{2}\left(-n_3+t+\sqrt{-2n_3^2+2s+2n_3t-t^2}\right)
$$
and
$$
n_1=\frac{1}{2}\left(-n_3+t+\sqrt{-2n_3^2+2s+2n_3t-t^2}\right)
$$
$$
n_2=\frac{1}{2}\left(-n_3+t-\sqrt{-2n_3^2+2s+2n_3t-t^2}\right)
$$
Hence if we set
$$
A(t,s)=\sum_{
\begin{array}{cc}
n_1+n_2+n_3=t\\
n_1^2+n_2^2+n_3^2=s\\
\end{array}
}1,
$$
then, if the discriminant of $n_1$ and $n_2$: $-2n_3^2+2s+2n_3t-t^2$ is zero possess one double root $(n_1=n_2)$. Otherwise if $-2n_3^2+2s+2n_3t-t^2=m^2$, $m\in\textbf{Z}^{*}$ two different. Hence if $s'=\left[\sqrt{s}\right]$ is the floor of the square root of $s\geq 0$, we get
$$
A(t,s)=\sum_{
\begin{array}{cc}
-s'\leq n_3\leq s'\\
-3n_3^2+2s+2n_3t-t^2=0
\end{array}
}1+
$$
$$
+2\sum_{
\begin{array}{cc}
-s'\leq n_3\leq s'\\
-3n_3^2+2s+2n_3t-t^2\neq 0
\end{array}
}X_{\textbf{Z}}\left(\frac{1}{2}\left(-n_3+t+\sqrt{-3n_3^2+2s+2n_3t-t^2}\right)\right),
$$
where $X_{\textbf{Z}}(n)$ is the characteristic function on integers.
By this way we get if $r_0(t,2s)$ is the number of representations of $2s$ in the form
$$
3x^2-2xt+t^2
$$
and $r_1(t,2s)$ the number of representations of $2s$ in the form
$$
3x^2-2xt+m^2+t^2,
$$
then $r(t,2s)=r_0(t,2s)+2r_1(t,2s)=A(t,s)$ is the Fourier coefficient of
$$
\Theta(q)=\sum^{\infty}_{t=-\infty}\sum^{\infty}_{s=0}A(t,s)q^{(t+1)^2+s}\textrm{, }|q|<1
$$
If $s-t=p$, then
$$
2s=3x^2-2xt+m^2+t^2\Leftrightarrow 3x^2-2tx+t^2-2t+m^2=2p.\tag 1
$$
We want to show that the above last equation have always integer solutions for every even non negative integer $p\geq0$ and for $p$ odd positive none. This will enable us to conclude, that the odd Fourier coefficients of $\Theta(q)$ are always occur and are non zero iff $p$ is even. For $p$ odd we have no representations. For this we define:
$$
\phi(q)=\sum^{\infty}_{n,t=-\infty}q^{3n^2-2tn+t^2-2t}
$$
and assume the transformation $n\rightarrow an+bt$, $t\rightarrow c n+d t$, where $a=-1$, $b=1$, $c=1$, $d=-2$. Then $ad-bc=1$ and
$$
\phi(q)=2\sum^{\infty}_{n,t=-\infty}q^{n^2-2n+8 t^2+4t}.
$$
Hence
$$
\phi(q)=2q^{-1}\left(\sum^{\infty}_{n=-\infty}q^{n^2-2n+1}\right)\left(\sum^{\infty}_{t=-\infty}q^{8t^2+4t}\right)=
$$
$$
=2q^{-1}\theta_3(q)\sum^{\infty}_{n=-\infty}q^{8n^2+4n},
$$
where $\theta_3(q)=\sum^{\infty}_{n=-\infty}q^{n^2}$, $|q|<1$. Hence (1) have analog:
$$
\phi(q)\theta_3(q)=\sum_{n,k,m\in\textbf{Z}}q^{3n^2-2nk+k^2-k+m^2}=2q^{-1}\theta_3(q)^2\sum^{\infty}_{n=-\infty}q^{8n^2+4n}=
$$
$$
=2q^{-1}\theta_3(q)^2\phi_e(q^2),
$$
where $\phi_e(q)=\sum^{\infty}_{n=-\infty}q^{4n^2+2n}$. But
$$
\theta_2(q)=q^{1/4}\sum^{\infty}_{n=-\infty}q^{n^2+n}=q^{1/4}\sum^{\infty}_{n=-\infty}q^{4n^2+2n}+\sum^{\infty}_{n=-\infty}q^{(2n+1+1/2)^2}
$$
and
$$
\sum^{\infty}_{n=-\infty}q^{(2n+1+1/2)^2}=\sum^{\infty}_{n=-\infty}q^{(2n+3/2-2)^2}=
$$
$$
\sum^{\infty}_{n=-\infty}q^{(2n-1/2)^2}=q^{1/4}\sum^{\infty}_{n=-\infty}q^{4n^2-2n}.
$$
Hence
$$
\theta_2(q)=2q^{1/4}\sum^{\infty}_{n=-\infty}q^{4n^2+2n}
$$
and therefore
$$
\phi(q)\theta_3(q)=q^{-3/2}\theta_3(q)^2\theta_2(q^2).\tag 2
$$
Hence is equivalent to show that
$$
x^2+y^2+2z^2+2z=2p+1,\tag 3
$$
have always solutions when $p$ is even non negative integer and none when $p$ positive odd. But indeed: $2z^2+2z\equiv 0(4)$ and when $p$ is odd we have no solutions since no integer of the form $4n+3$ have representation as a sum of two squares. Hence we left with case $p$ even.
The function $f(z)=\theta_3(q)^2\theta_2(q^2)$, $q=e(z)=e^{2\pi i z}$, $Im(z)>0$ is the associated theta function of (3) and is a modular form of weight $3/2$ in $\Gamma(8)$, i.e. it holds
$$
f\left(\frac{az+b}{cz+d}\right)=\epsilon_{c,d}(cz+d)^{3/2}f(z),
$$
when $Im(z)>0$ and $a,b,c,d$ integers with $ad-bc=1$, $a,d\equiv 1(8)$, $c,b\equiv 0(8)$. The function $\epsilon_{c,d}$ is such that $\epsilon_{c,d}^4=1$. But I don't have much experience to go further.
