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The desired integral is $I(x)$, defined by $$I(x)=e^{N^2r\frac{1}{k}x^2}J(x),\;\;J(x)=\mathop{\idotsint\limits_{\mathbb{R}^k}}_{\sum_{j=1}^{k}y_j=x} e^{-N^2r\sum_{j=1}^{k}y_j^2} dy_1\dots dy_{k}.$$ Fourier transform $J(x)$, $$F(q)=\int_{-\infty}^\infty e^{iqx}J(x)\,dx=\mathop{\idotsint} e^{-N^2r\sum_{j=1}^{k}y_j^2}e^{\sum_{j=1}^k qy_j} dy_1\dots dy_{k}=N^{-k}(\pi/r)^{k/2}e^{-kq^2/(4N^2r)}.$$ Now Fourier transform back from $F(q)$ to $J(x)$ and you're done: $$J(x)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-iqx}F(q)\,dq=k^{-1/2} N^{1-k} (r/\pi)^{(1-k)/2} e^{-N^2 r x^2/k}$$ $$\Rightarrow I(x)=k^{-1/2} N^{1-k} (r/\pi)^{(1-k)/2}.$$ This differs from both of the answers in the OP.