The answer is negative if the complex torus is not algebraic (or equivalently, not Moishezon). More generally, if $U$ is the analytification of a separated scheme of finite type over $\mathbf{C}$ and $Y$ is a proper complex-analytic space admitting an open immersion $j:U \hookrightarrow Y$ onto the complement of a nowhere-dense analytic set then $Y$ must be Moishzeon. However, even if $Y$ is a manifold, without knowing the geometry of the complement of $Y-U$ in $Y$ perhaps the restriction map $M_Y\rightarrow j_{\ast}(M_U)$ with sheaves of meromorphic functions is not an equality on global sections (i.e., the Levi extension theorem may not apply). Thus, to prove that $Y$ must be Moishezon it is necessary to use more serious input from complex analysis as follows.

We may assume $Y$ is reduced and irreducible (by the very definition of "Moishzeon") and may shrink $U$ to be affine (if it weren't already quasi-projective) and even smooth (by reducedness). The normalization $Y'$ of $Y$ is connected and $Y' \rightarrow Y$ is an isomorphism over $U$, and $M(Y) = M(Y')$ by the theory of analytic normalization, so we may replace $Y$ with $Y'$ to arrange that $Y$ is normal. By resolution of singularities, $U$ is the complement of a strict normal crossings divisor in a connected projective manifold $X$. Borel's extension theorem extends $U\rightarrow Y$ to $\nu:X\rightarrow Y$ by normality of $Y$. But $M_Y\rightarrow\nu_{\ast}(M_X)$ is an isomorphism by 10.6.2 in "Coherent Analytic Sheaves" (or really Remark 8.1.3 in loc. cit.), which rests crucially on Grauert's coherence theorem, so $M_Y(Y) = M_X(X)$ has transcendence degree $\dim X = \dim Y$ over $\mathbf{C}$. Hence, $Y$ is Moishezon.